What are the solutions to the equation s^2-22s=31 ?

The solutions to the equation s^2-22s=31 are s=11+2sqrt(38),s=11-2sqrt(38)

Find the zeros of s^2-22s=31

The zeros of s^2-22s=31 are s=11+2sqrt(38),s=11-2sqrt(38)

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s^{2} - 22 s = 31

s = 11 + 238, s = 11 - 238

Decimal Notation

s = 23.32882 \dots, s = - 1.32882 \dots

Solution steps

s^{2} - 22 s = 31

Move

31

to the left side

s^{2} - 22 s - 31 = 0

Solve with the quadratic formula

s_{1, 2} = \frac{- ( - 22 ) \pm ( - 22 ) ^{2} - 4 \cdot 1 \cdot ( - 31 )}{2 \cdot 1}

(- 22)^{2} - 4 \cdot 1 \cdot (- 31) = 438

s_{1, 2} = \frac{- ( - 22 ) \pm 4 38}{2 \cdot 1}

Separate the solutions

s_{1} = \frac{- ( - 22 ) + 4 38}{2 \cdot 1}, s_{2} = \frac{- ( - 22 ) - 4 38}{2 \cdot 1}

s = \frac{- ( - 22 ) + 4 38}{2 \cdot 1} : 11 + 238

s = \frac{- ( - 22 ) - 4 38}{2 \cdot 1} : 11 - 238

The solutions to the quadratic equation are:

s = 11 + 238, s = 11 - 238

a (x) = - \frac{8}{7} x^{2} + 240 x

7 x^{2} + 32 x - 15 = 0

- (x + 1) (3 x + 2) = 0

2 x^{2} - 23 x = - 7

3 y^{2} + 2 y - 5 = 0

What are the solutions to the equation s^2-22s=31 ?
The solutions to the equation s^2-22s=31 are s=11+2sqrt(38),s=11-2sqrt(38)
Find the zeros of s^2-22s=31
The zeros of s^2-22s=31 are s=11+2sqrt(38),s=11-2sqrt(38)