What are the solutions to the equation s^2+24s=-35 ?

The solutions to the equation s^2+24s=-35 are s=-12+sqrt(109),s=-12-sqrt(109)

Find the zeros of s^2+24s=-35

The zeros of s^2+24s=-35 are s=-12+sqrt(109),s=-12-sqrt(109)

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s^{2} + 24 s = - 35

s = - 12 + 109, s = - 12 - 109

Decimal Notation

s = - 1.55969 \dots, s = - 22.44030 \dots

Solution steps

s^{2} + 24 s = - 35

Move

35

to the left side

s^{2} + 24 s + 35 = 0

Solve with the quadratic formula

s_{1, 2} = \frac{- 24 \pm 2 4 ^{2} - 4 \cdot 1 \cdot 35}{2 \cdot 1}

2 4^{2} - 4 \cdot 1 \cdot 35 = 2109

s_{1, 2} = \frac{- 24 \pm 2 109}{2 \cdot 1}

Separate the solutions

s_{1} = \frac{- 24 + 2 109}{2 \cdot 1}, s_{2} = \frac{- 24 - 2 109}{2 \cdot 1}

s = \frac{- 24 + 2 109}{2 \cdot 1} : - 12 + 109

s = \frac{- 24 - 2 109}{2 \cdot 1} : - 12 - 109

The solutions to the quadratic equation are:

s = - 12 + 109, s = - 12 - 109

lo g_{\frac{1}{9}} (81)

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

4 x^{2} - 5 x + 10 = 0

\frac{2 ^{5}}{3}

- \frac{113}{( 0 + 6 ) ^{2}}

What are the solutions to the equation s^2+24s=-35 ?
The solutions to the equation s^2+24s=-35 are s=-12+sqrt(109),s=-12-sqrt(109)
Find the zeros of s^2+24s=-35
The zeros of s^2+24s=-35 are s=-12+sqrt(109),s=-12-sqrt(109)