Question

2x2+3x-5=0 alguemmm me ajudaaaa (braskara)​

Answer 1

Resposta:

Explicação passo-a-passo:

$\displaystyle Aplicando~a~f\'{o}rmula~de~Bhaskara~para~2x^{2}+3x-5=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=2{;}~b=3~e~c=-5\\\\\Delta=(b)^{2}-4(a)(c)=(3)^{2}-4(2)(-5)=9-(-40)=49\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(3)-\sqrt{49}}{2(2)}=\frac{-3-7}{4}=\frac{-10\div2}{4\div2}=-\frac{5}{2}\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(3)+\sqrt{49}}{2(2)}=\frac{-3+7}{4}=\frac{4}{4}=1\\\\\\S=\{-\frac{5}{2}\,~1\}$

Answer 2

Resposta:

• Usar a fórmula de delta