Question

решить уравнение:
$2x^2-5x-3=0$

Answer 1

$\displaystyle \tt 2x^2-5x-3=0\\\\ \displaystyle \tt D=(-5)^2-4\cdot2\cdot(-3)=25+24=49\\\\ \displaystyle \tt \sqrt{D}=\sqrt{49}=7\\\\ \displaystyle \tt x_1=\frac{5+7}{2\cdot2}=\frac{12}{4}=3\\\\\displaystyle \tt x_2=\frac{5-7}{2\cdot2}=\frac{-2}{4}=-\frac{1}{2}=-0,5$

Answer 2

$2 {x}^{2} - 5x - 3 = 0 \\ D = 25 - 4 \times 2 \times ( - 3) = 25 + 24 = 49 = {7}^{2} \\ x_1 = \frac{5 + 7}{4} = \frac{12}{4} = 3 \\ x_2= \frac{5 - 7}{4} = \frac{ - 2}{4} = - 0.5$