What are the solutions to the equation 10x^2+19x+6=(5x+a)(bx+c) ?

The solutions to the equation 10x^2+19x+6=(5x+a)(bx+c) are

Find the zeros of 10x^2+19x+6=(5x+a)(bx+c)

The zeros of 10x^2+19x+6=(5x+a)(bx+c) are

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10 x^{2} + 19 x + 6 = (5 x + a) (b x + c)

x = \frac{- 19 + ab + 5 c + a ^{2} b ^{2} + 40 a c - 38 ab - 10 ab c + 120 b + 25 c ^{2} + 121 - 190 c}{2 ( 10 - 5 b )}, x = \frac{- 19 + ab + 5 c - a ^{2} b ^{2} + 40 a c - 38 ab - 10 ab c + 120 b + 25 c ^{2} + 121 - 190 c}{2 ( 10 - 5 b )}; b \neq = 2

Solution steps

10 x^{2} + 19 x + 6 = (5 x + a) (b x + c)

Expand

(5 x + a) (b x + c) : 5 b x^{2} + 5 c x + ab x + a c

10 x^{2} + 19 x + 6 = 5 b x^{2} + 5 c x + ab x + a c

Subtract

ab x + a c

from both sides

10 x^{2} + 19 x + 6 - (ab x + a c) = 5 b x^{2} + 5 c x + ab x + a c - (ab x + a c)

Simplify

10 x^{2} + 19 x + 6 - ab x - a c = 5 b x^{2} + 5 c x

Move

5 c x

to the left side

10 x^{2} + 19 x + 6 - ab x - a c - 5 c x = 5 b x^{2}

Move

5 b x^{2}

to the left side

10 x^{2} + 19 x + 6 - ab x - a c - 5 c x - 5 b x^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

(10 - 5 b) x^{2} + (19 - ab - 5 c) x + 6 - a c = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( 19 - ab - 5 c ) \pm ( 19 - ab - 5 c ) ^{2} - 4 ( 10 - 5 b ) ( 6 - a c )}{2 ( 10 - 5 b )}

Simplify

(19 - ab - 5 c)^{2} - 4 (10 - 5 b) (6 - a c) : a^{2} b^{2} + 40 a c - 38 ab - 10 ab c + 120 b + 25 c^{2} + 121 - 190 c

x_{1, 2} = \frac{- ( 19 - ab - 5 c ) \pm a ^{2} b ^{2} + 40 a c - 38 ab - 10 ab c + 120 b + 25 c ^{2} + 121 - 190 c}{2 ( 10 - 5 b )}; b \neq = 2

Separate the solutions

x_{1} = \frac{- ( 19 - ab - 5 c ) + a ^{2} b ^{2} + 40 a c - 38 ab - 10 ab c + 120 b + 25 c ^{2} + 121 - 190 c}{2 ( 10 - 5 b )}, x_{2} = \frac{- ( 19 - ab - 5 c ) - a ^{2} b ^{2} + 40 a c - 38 ab - 10 ab c + 120 b + 25 c ^{2} + 121 - 190 c}{2 ( 10 - 5 b )}

x = \frac{- ( 19 - ab - 5 c ) + a ^{2} b ^{2} + 40 a c - 38 ab - 10 ab c + 120 b + 25 c ^{2} + 121 - 190 c}{2 ( 10 - 5 b )} : \frac{- 19 + ab + 5 c + a ^{2} b ^{2} + 40 a c - 38 ab - 10 ab c + 120 b + 25 c ^{2} + 121 - 190 c}{2 ( 10 - 5 b )}

x = \frac{- ( 19 - ab - 5 c ) - a ^{2} b ^{2} + 40 a c - 38 ab - 10 ab c + 120 b + 25 c ^{2} + 121 - 190 c}{2 ( 10 - 5 b )} : \frac{- 19 + ab + 5 c - a ^{2} b ^{2} + 40 a c - 38 ab - 10 ab c + 120 b + 25 c ^{2} + 121 - 190 c}{2 ( 10 - 5 b )}

The solutions to the quadratic equation are:

x = \frac{- 19 + ab + 5 c + a ^{2} b ^{2} + 40 a c - 38 ab - 10 ab c + 120 b + 25 c ^{2} + 121 - 190 c}{2 ( 10 - 5 b )}, x = \frac{- 19 + ab + 5 c - a ^{2} b ^{2} + 40 a c - 38 ab - 10 ab c + 120 b + 25 c ^{2} + 121 - 190 c}{2 ( 10 - 5 b )}; b \neq = 2

- 4 x^{2} + 8 x - 4 = 0

1 - 1.2 z + 0.85 \cdot z^{2} = 0

15 + 2 x^{2} = 177

- 3 x^{2} - 4 = 0

0.000006 x^{2} - 0.02 x + 25 = 15

What are the solutions to the equation 10x^2+19x+6=(5x+a)(bx+c) ?
The solutions to the equation 10x^2+19x+6=(5x+a)(bx+c) are
Find the zeros of 10x^2+19x+6=(5x+a)(bx+c)
The zeros of 10x^2+19x+6=(5x+a)(bx+c) are