Question

${ - x}^{2}-10x + 20 = 0$
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Answer 1

Resolvendo por fórmula de Bháskara, temos:

$-x^2-10x+20=0 \\ \hookrightarrow -x^2-10x+20=0 * (-1) \\ \hookrightarrow \boxed{x^2+10x-20=0} \\\\ \Delta = 10^2-4*1*(-20) \\ \Delta = 100+80 \\ \Delta = 180 \\\\ x = \frac{-b\pm\sqrt{\Delta}}{2a} \\ x = \frac{-10\pm\sqrt{180}}{2*1} \\ x' = \frac{-10+\sqrt{2^2*3^2*5}}{2} = \frac{-10+2*3\sqrt{5}}{2} = \frac{-10+6\sqrt{5} }{2} = \frac{2*(-5+3\sqrt{5})}{2} = \boxed{-5+3\sqrt{5}}$

$x'' = \frac{-10-\sqrt{2^2*3^2*5}}{2} = \frac{-10-2*3\sqrt{5}}{2} = \frac{-10-6\sqrt{5}}{2} = \frac{2*(-5-3\sqrt{5})}{2} = \boxed{-5-3\sqrt{5}}$