What are the solutions to the equation x^2-5x=21 ?

The solutions to the equation x^2-5x=21 are x=(5+sqrt(109))/2 ,x=(5-sqrt(109))/2

Find the zeros of x^2-5x=21

The zeros of x^2-5x=21 are x=(5+sqrt(109))/2 ,x=(5-sqrt(109))/2

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x^{2} - 5 x = 21

x = \frac{5 + 109}{2}, x = \frac{5 - 109}{2}

Decimal Notation

x = 7.72015 \dots, x = - 2.72015 \dots

Solution steps

x^{2} - 5 x = 21

Move

21

to the left side

x^{2} - 5 x - 21 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot ( - 21 )}{2 \cdot 1}

(- 5)^{2} - 4 \cdot 1 \cdot (- 21) = 109

x_{1, 2} = \frac{- ( - 5 ) \pm 109}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 109}{2 \cdot 1}, x_{2} = \frac{- ( - 5 ) - 109}{2 \cdot 1}

x = \frac{- ( - 5 ) + 109}{2 \cdot 1} : \frac{5 + 109}{2}

x = \frac{- ( - 5 ) - 109}{2 \cdot 1} : \frac{5 - 109}{2}

The solutions to the quadratic equation are:

x = \frac{5 + 109}{2}, x = \frac{5 - 109}{2}

h, (2 - t) ((5 - h - t) (5 + h - t)) + h^{2} = 0

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

x^{2} - 5 x = - 7

(37.15 - 1 x) \cdot (10.19 - 0.5 x) - (15.12)^{2} = 0

- 0.23 x^{2} + 3.68 x + 8.21 = 0

What are the solutions to the equation x^2-5x=21 ?
The solutions to the equation x^2-5x=21 are x=(5+sqrt(109))/2 ,x=(5-sqrt(109))/2
Find the zeros of x^2-5x=21
The zeros of x^2-5x=21 are x=(5+sqrt(109))/2 ,x=(5-sqrt(109))/2