What are the solutions to the equation 32x-4=4x^2+60 ?

The solutions to the equation 32x-4=4x^2+60 are x=4

Find the zeros of 32x-4=4x^2+60

The zeros of 32x-4=4x^2+60 are x=4

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32 x - 4 = 4 x^{2} + 60

x = 4

Solution steps

32 x - 4 = 4 x^{2} + 60

Switch sides

4 x^{2} + 60 = 32 x - 4

Move

4

to the left side

4 x^{2} + 64 = 32 x

Move

32 x

to the left side

4 x^{2} + 64 - 32 x = 0

Write in the standard form

a x^{2} + b x + c = 0

4 x^{2} - 32 x + 64 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 32 ) \pm ( - 32 ) ^{2} - 4 \cdot 4 \cdot 64}{2 \cdot 4}

(- 32)^{2} - 4 \cdot 4 \cdot 64 = 0

x_{1, 2} = \frac{- ( - 32 ) \pm 0}{2 \cdot 4}

x = \frac{- ( - 32 )}{2 \cdot 4}

\frac{- ( - 32 )}{2 \cdot 4} = 4

x = 4

The solution to the quadratic equation is:

x = 4

- x^{2} + 10 x + 11 = 0

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

3 x^{2} + 2 x

y^{2} + 3 y - 7 = 0

x^{2} + 60 x + 1000 = 0

What are the solutions to the equation 32x-4=4x^2+60 ?
The solutions to the equation 32x-4=4x^2+60 are x=4
Find the zeros of 32x-4=4x^2+60
The zeros of 32x-4=4x^2+60 are x=4