What are the solutions to the equation (x-5)(4x+2)=-3 ?

The solutions to the equation (x-5)(4x+2)=-3 are x=(9+sqrt(109))/4 ,x=(9-sqrt(109))/4

Find the zeros of (x-5)(4x+2)=-3

The zeros of (x-5)(4x+2)=-3 are x=(9+sqrt(109))/4 ,x=(9-sqrt(109))/4

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(x - 5) (4 x + 2) = - 3

x = \frac{9 + 109}{4}, x = \frac{9 - 109}{4}

Decimal Notation

x = 4.86007 \dots, x = - 0.36007 \dots

Solution steps

(x - 5) (4 x + 2) = - 3

Expand

(x - 5) (4 x + 2) : 4 x^{2} - 18 x - 10

4 x^{2} - 18 x - 10 = - 3

Move

3

to the left side

4 x^{2} - 18 x - 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 \cdot 4 ( - 7 )}{2 \cdot 4}

(- 18)^{2} - 4 \cdot 4 (- 7) = 2109

x_{1, 2} = \frac{- ( - 18 ) \pm 2 109}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 18 ) + 2 109}{2 \cdot 4}, x_{2} = \frac{- ( - 18 ) - 2 109}{2 \cdot 4}

x = \frac{- ( - 18 ) + 2 109}{2 \cdot 4} : \frac{9 + 109}{4}

x = \frac{- ( - 18 ) - 2 109}{2 \cdot 4} : \frac{9 - 109}{4}

The solutions to the quadratic equation are:

x = \frac{9 + 109}{4}, x = \frac{9 - 109}{4}

x, f = x^{2} + 4 y

0 = 8 - 2 x^{2}

12 x^{2} + 18 x = 0

0 = x^{2} + 2 x - 1

z^{2} = - 5

What are the solutions to the equation (x-5)(4x+2)=-3 ?
The solutions to the equation (x-5)(4x+2)=-3 are x=(9+sqrt(109))/4 ,x=(9-sqrt(109))/4
Find the zeros of (x-5)(4x+2)=-3
The zeros of (x-5)(4x+2)=-3 are x=(9+sqrt(109))/4 ,x=(9-sqrt(109))/4