What are the solutions to the equation (x^2-4x)+(3x-2)=28 ?

The solutions to the equation (x^2-4x)+(3x-2)=28 are x=6,x=-5

Find the zeros of (x^2-4x)+(3x-2)=28

The zeros of (x^2-4x)+(3x-2)=28 are x=6,x=-5

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(x^{2} - 4 x) + (3 x - 2) = 28

x = 6, x = - 5

Solution steps

(x^{2} - 4 x) + (3 x - 2) = 28

Refine

x^{2} - x - 2 = 28

Move

28

to the left side

x^{2} - x - 30 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 30 )}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot (- 30) = 11

x_{1, 2} = \frac{- ( - 1 ) \pm 11}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 11}{2 \cdot 1}, x_{2} = \frac{- ( - 1 ) - 11}{2 \cdot 1}

x = \frac{- ( - 1 ) + 11}{2 \cdot 1} : 6

x = \frac{- ( - 1 ) - 11}{2 \cdot 1} : - 5

The solutions to the quadratic equation are:

x = 6, x = - 5

0 = - 16 t^{2} + 100

s^{2} + 8 s + 25 + k s^{2} + 8 k s + 12 k = 0

(2 x + 1) (5 x - 3) = 0

4 a^{2} - 81 = 0

5 x^{2} + 4 = 21 x

What are the solutions to the equation (x^2-4x)+(3x-2)=28 ?
The solutions to the equation (x^2-4x)+(3x-2)=28 are x=6,x=-5
Find the zeros of (x^2-4x)+(3x-2)=28
The zeros of (x^2-4x)+(3x-2)=28 are x=6,x=-5