Question

Solve for \(x\) in the quadratic equation \(x^{2}+10 x+8=0\)

Answer 1

\(x=-0,88\) or \(x=-9,12\)

 

Explanation:

\(\begin{aligned} x^{2} &+10 x+8=0 \\ x &=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\ &=\frac{-10 \pm \sqrt{10^{2}-4(1)(8)}}{2(1)} \\ &=\frac{-10 \pm \sqrt{68}}{2} \\ x &=-0,88 \text { or } x=-9,12 \end{aligned}\)