What are the solutions to the equation quadraticformula (2x+1)(2x-2)=88 ?

The solutions to the equation quadraticformula (2x+1)(2x-2)=88 are x=5,x=-9/2

Find the zeros of quadraticformula (2x+1)(2x-2)=88

The zeros of quadraticformula (2x+1)(2x-2)=88 are x=5,x=-9/2

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(2 x + 1) (2 x - 2) = 88

x = 5, x = - \frac{9}{2}

Decimal Notation

x = 5, x = - 4.5

Solution steps

(2 x + 1) (2 x - 2) = 88

Expand

(2 x + 1) (2 x - 2) : 4 x^{2} - 2 x - 2

4 x^{2} - 2 x - 2 = 88

Move

88

to the left side

4 x^{2} - 2 x - 90 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 4 ( - 90 )}{2 \cdot 4}

(- 2)^{2} - 4 \cdot 4 (- 90) = 38

x_{1, 2} = \frac{- ( - 2 ) \pm 38}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 38}{2 \cdot 4}, x_{2} = \frac{- ( - 2 ) - 38}{2 \cdot 4}

x = \frac{- ( - 2 ) + 38}{2 \cdot 4} : 5

x = \frac{- ( - 2 ) - 38}{2 \cdot 4} : - \frac{9}{2}

The solutions to the quadratic equation are:

x = 5, x = - \frac{9}{2}

7 x^{2} = 9

4 x^{2} - 11 x + 5 = 0

15 t - 1.86 t^{2} - 25 = 0

x^{2} - 4 x = 2 x - x^{2}

56 = - x^{2} + 15 x

What are the solutions to the equation quadraticformula (2x+1)(2x-2)=88 ?
The solutions to the equation quadraticformula (2x+1)(2x-2)=88 are x=5,x=-9/2
Find the zeros of quadraticformula (2x+1)(2x-2)=88
The zeros of quadraticformula (2x+1)(2x-2)=88 are x=5,x=-9/2