Question

Solução de x2 +14x +40 = 0

Answer 1

Resposta:

$S = \{ - 10; \: - 4 \}$

Explicação passo-a-passo:

${x}^{2} + 14x + 40 = 0 \\ \\ a = 1 \\ b = 14 \\ c = 40 \\ \\ \Delta = {b}^{2} - 4ac \\ \Delta = {14}^{2} - 4 \cdot1 \cdot40 \\ \Delta = 196 - 160 \\ \Delta = 36$

$x = \frac{ - b\pm \sqrt{\Delta} }{2a} \\ \\ x = \frac{ - 14\pm \sqrt{36} }{2\cdot1} \\ \\ x = \frac{ - 14\pm6}{2}$

$x_{1}= \frac{ - 14 + 6}{2} = - \frac{8}{2} = \textbf{ - 4} \\ \\ x_{2} = \frac{ - 14 - 6}{2} = - \frac{20}{2} = \textbf{ - 10}$

Espero ter ajudado! :)

Answer 2

Resposta:

$Aplicando~a~f\'ormula~de~Bhaskara~para~x^{2}+14x+40=0~~\\e~comparando~com~ f(x)=(a)x^{2}+(b)x+(c),~temos~a=1{;}~b=14~e~c=40\\\\\Delta=(b)^{2}-4(a)(c)=(14)^{2}-4(1)(40)=196-(160)=36\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(14)-\sqrt{36}}{2(1)}=\frac{-14-6}{2}=\frac{-20}{2}=-10\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(14)+\sqrt{36}}{2(1)}=\frac{-14+6}{2}=\frac{-8}{2}=-4\\\\S=\{-10,~-4\}$