Question

Resolva a Equação do 2o grau2x2 + x – 3 = 0.

Answer 1

$2x^2+x-3=0\\\\x_{1,\:2}=\frac{-1\pm \sqrt{1^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}\\\\x_{1,\:2}=\frac{-1\pm \:5}{2\cdot \:2}\\\\x_1=\frac{-1+5}{2\cdot \:2},\:x_2=\frac{-1-5}{2\cdot \:2}\\\\\frac{-1+5}{2\cdot \:2}\\\\=\frac{4}{2\cdot \:2}\\\\=\frac{4}{4}\\\\=1\\\\\frac{-1-5}{2\cdot \:2}\\\\=\frac{-6}{2\cdot \:2}\\\\=\frac{-6}{4}\\\\=-\frac{6}{4}\\\\=-\frac{3}{2}$

$\mathrm{As\:solucoes\:para\:a\:equacao\:de\:segundo\:grau\:sao:\:}$

$x=1,\:x=-\frac{3}{2}$

Answer 2

Resposta:

1 ou -3/2

Explicação passo-a-passo:

2x^2 + x - 3 = 0

utilizando a fórmula de Bhaskara ($\frac{-b +- \sqrt{b^{2} - 4ac } }{2a}$ )

$\frac{-1 +- \sqrt{1 - 4(2)(-3)} } {2*2}$

$\frac{-1+-\sqrt{25} }{4}$

$\frac{-1+-5}{4}$

Nesse ponto é necessário testar os dois casos, assim:

$\frac{-1+5}{4}= \frac{4}{4} = 1$

ou

$\frac{-1-5}{4} = \frac{-6}{4} = \frac{-3}{2}$