What are the solutions to the equation x^2-3/4 x=5 ?

The solutions to the equation x^2-3/4 x=5 are x=(3+sqrt(329))/8 ,x=(3-sqrt(329))/8

Find the zeros of x^2-3/4 x=5

The zeros of x^2-3/4 x=5 are x=(3+sqrt(329))/8 ,x=(3-sqrt(329))/8

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x^{2} - \frac{3}{4} x = 5

x = \frac{3 + 329}{8}, x = \frac{3 - 329}{8}

Decimal Notation

x = 2.64229 \dots, x = - 1.89229 \dots

Solution steps

x^{2} - \frac{3}{4} x = 5

Multiply both sides by

4

4 x^{2} - 3 x = 20

Move

20

to the left side

4 x^{2} - 3 x - 20 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 4 ( - 20 )}{2 \cdot 4}

(- 3)^{2} - 4 \cdot 4 (- 20) = 329

x_{1, 2} = \frac{- ( - 3 ) \pm 329}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 329}{2 \cdot 4}, x_{2} = \frac{- ( - 3 ) - 329}{2 \cdot 4}

x = \frac{- ( - 3 ) + 329}{2 \cdot 4} : \frac{3 + 329}{8}

x = \frac{- ( - 3 ) - 329}{2 \cdot 4} : \frac{3 - 329}{8}

The solutions to the quadratic equation are:

x = \frac{3 + 329}{8}, x = \frac{3 - 329}{8}

x^{2} - 6 x - 16 < 0

\frac{1}{4} 2

(a + b)^{3}

2 x^{2} + 6 x = - 9

\frac{9}{36}

What are the solutions to the equation x^2-3/4 x=5 ?
The solutions to the equation x^2-3/4 x=5 are x=(3+sqrt(329))/8 ,x=(3-sqrt(329))/8
Find the zeros of x^2-3/4 x=5
The zeros of x^2-3/4 x=5 are x=(3+sqrt(329))/8 ,x=(3-sqrt(329))/8