What are the solutions to the equation 4x^2-20x+16=-8 ?

The solutions to the equation 4x^2-20x+16=-8 are x=3,x=2

Find the zeros of 4x^2-20x+16=-8

The zeros of 4x^2-20x+16=-8 are x=3,x=2

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4 x^{2} - 20 x + 16 = - 8

x = 3, x = 2

Solution steps

4 x^{2} - 20 x + 16 = - 8

Move

8

to the left side

4 x^{2} - 20 x + 24 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 \cdot 4 \cdot 24}{2 \cdot 4}

(- 20)^{2} - 4 \cdot 4 \cdot 24 = 4

x_{1, 2} = \frac{- ( - 20 ) \pm 4}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 20 ) + 4}{2 \cdot 4}, x_{2} = \frac{- ( - 20 ) - 4}{2 \cdot 4}

x = \frac{- ( - 20 ) + 4}{2 \cdot 4} : 3

x = \frac{- ( - 20 ) - 4}{2 \cdot 4} : 2

The solutions to the quadratic equation are:

x = 3, x = 2

x^{2} - 14 x = - 24

4 x^{2} - 8 x - 2

4 x^{2} - 8 x + 1

x^{2} + 5 x + 6 = 12

5 v^{2} + 29 v + 20 = 0

What are the solutions to the equation 4x^2-20x+16=-8 ?
The solutions to the equation 4x^2-20x+16=-8 are x=3,x=2
Find the zeros of 4x^2-20x+16=-8
The zeros of 4x^2-20x+16=-8 are x=3,x=2