What are the solutions to the equation 4x^2+236=64x ?

The solutions to the equation 4x^2+236=64x are x=8+sqrt(5),x=8-sqrt(5)

Find the zeros of 4x^2+236=64x

The zeros of 4x^2+236=64x are x=8+sqrt(5),x=8-sqrt(5)

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4 x^{2} + 236 = 64 x

x = 8 + 5, x = 8 - 5

Decimal Notation

x = 10.23606 \dots, x = 5.76393 \dots

Solution steps

4 x^{2} + 236 = 64 x

Move

64 x

to the left side

4 x^{2} + 236 - 64 x = 0

Write in the standard form

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

4 x^{2} - 64 x + 236 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 64 ) \pm ( - 64 ) ^{2} - 4 \cdot 4 \cdot 236}{2 \cdot 4}

(- 64)^{2} - 4 \cdot 4 \cdot 236 = 85

x_{1, 2} = \frac{- ( - 64 ) \pm 8 5}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 64 ) + 8 5}{2 \cdot 4}, x_{2} = \frac{- ( - 64 ) - 8 5}{2 \cdot 4}

x = \frac{- ( - 64 ) + 8 5}{2 \cdot 4} : 8 + 5

x = \frac{- ( - 64 ) - 8 5}{2 \cdot 4} : 8 - 5

The solutions to the quadratic equation are:

x = 8 + 5, x = 8 - 5

x^{2} + 14 = 30

x^{2} - 6 x + 15 = 3 (x - 1)

24 + x^{2} = 10 x

x, y = x - x^{2}

z^{2} + 5 z - 24 = 0

What are the solutions to the equation 4x^2+236=64x ?
The solutions to the equation 4x^2+236=64x are x=8+sqrt(5),x=8-sqrt(5)
Find the zeros of 4x^2+236=64x
The zeros of 4x^2+236=64x are x=8+sqrt(5),x=8-sqrt(5)