Solve for \(\mathrm{x}: 2 x^{2}-3 x-5=0\) [solved]

Question

Solve for \(\mathrm{x}: 2 x^{2}-3 x-5=0\)

Answer 1

Solve for \(x: 2 x^{2}-3 x-5=0\)
\[
\begin{aligned}
&a=2, b=-3, c=-5 \\
&x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\
&x=\frac{-(-3) \pm \sqrt{(-3)^{2}-4(2)(-5)}}{2(2)} \\
&x=\frac{3 \pm \sqrt{(-3)^{2}-4(2)(-5)}}{2(2)} \\
&x=\frac{3 \pm \sqrt{49}}{4} \\
&x=\frac{3 \pm 7}{4} \\
&x=\frac{3+7}{4}, x=\frac{3-7}{4} \\
&x=2.5, x=-1 \\
&\text { In } \frac{-b \pm \sqrt{\Delta}}{2 a}, \Delta=b^{2}-4 a c . \\
&\text { If } \Delta>0, \text { there are two solutions. } \\
&\text { If } \Delta=0, \text { there is only one solution. } \\
&\text { If } \Delta<0, \text { there are no real solutions. }
\end{aligned}
\]