What are the solutions to the equation 4(x^2-4)-5(3+2x)=-7 ?

The solutions to the equation 4(x^2-4)-5(3+2x)=-7 are x=4,x=-3/2

Find the zeros of 4(x^2-4)-5(3+2x)=-7

The zeros of 4(x^2-4)-5(3+2x)=-7 are x=4,x=-3/2

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4 (x^{2} - 4) - 5 (3 + 2 x) = - 7

x = 4, x = - \frac{3}{2}

Decimal Notation

x = 4, x = - 1.5

Solution steps

4 (x^{2} - 4) - 5 (3 + 2 x) = - 7

Expand

4 (x^{2} - 4) - 5 (3 + 2 x) : 4 x^{2} - 10 x - 31

4 x^{2} - 10 x - 31 = - 7

Move

7

to the left side

4 x^{2} - 10 x - 24 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 4 ( - 24 )}{2 \cdot 4}

(- 10)^{2} - 4 \cdot 4 (- 24) = 22

x_{1, 2} = \frac{- ( - 10 ) \pm 22}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 10 ) + 22}{2 \cdot 4}, x_{2} = \frac{- ( - 10 ) - 22}{2 \cdot 4}

x = \frac{- ( - 10 ) + 22}{2 \cdot 4} : 4

x = \frac{- ( - 10 ) - 22}{2 \cdot 4} : - \frac{3}{2}

The solutions to the quadratic equation are:

x = 4, x = - \frac{3}{2}

(x - 7) (x + 4) = 0

3 (2 x - 5)^{2} - 15 = 48

- 3 x^{2} - 3 x - 9 = - 2

2 x^{2} + 10 x + 4 = 0

4 x^{2} + 6 x + 3 = 0

What are the solutions to the equation 4(x^2-4)-5(3+2x)=-7 ?
The solutions to the equation 4(x^2-4)-5(3+2x)=-7 are x=4,x=-3/2
Find the zeros of 4(x^2-4)-5(3+2x)=-7
The zeros of 4(x^2-4)-5(3+2x)=-7 are x=4,x=-3/2