What are the solutions to the equation 4x^2+36=32x ?

The solutions to the equation 4x^2+36=32x are x=4+sqrt(7),x=4-sqrt(7)

Find the zeros of 4x^2+36=32x

The zeros of 4x^2+36=32x are x=4+sqrt(7),x=4-sqrt(7)

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4 x^{2} + 36 = 32 x

x = 4 + 7, x = 4 - 7

Decimal Notation

x = 6.64575 \dots, x = 1.35424 \dots

Solution steps

4 x^{2} + 36 = 32 x

Move

32 x

to the left side

4 x^{2} + 36 - 32 x = 0

Write in the standard form

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

4 x^{2} - 32 x + 36 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 32 ) \pm ( - 32 ) ^{2} - 4 \cdot 4 \cdot 36}{2 \cdot 4}

(- 32)^{2} - 4 \cdot 4 \cdot 36 = 87

x_{1, 2} = \frac{- ( - 32 ) \pm 8 7}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 32 ) + 8 7}{2 \cdot 4}, x_{2} = \frac{- ( - 32 ) - 8 7}{2 \cdot 4}

x = \frac{- ( - 32 ) + 8 7}{2 \cdot 4} : 4 + 7

x = \frac{- ( - 32 ) - 8 7}{2 \cdot 4} : 4 - 7

The solutions to the quadratic equation are:

x = 4 + 7, x = 4 - 7

81 x^{2} - 1 = 0

X^{2} - 6 X + 9 = 0

12 x^{2} - 15 x = 0

r^{2} = - 8 r + 65

0 = 3 x^{2} + 2 x - 1

What are the solutions to the equation 4x^2+36=32x ?
The solutions to the equation 4x^2+36=32x are x=4+sqrt(7),x=4-sqrt(7)
Find the zeros of 4x^2+36=32x
The zeros of 4x^2+36=32x are x=4+sqrt(7),x=4-sqrt(7)