Which are the solutions of the quadratic equation?

Mathematics

1 answer:

yarga [219]3 years ago

6 0

$\boxed{x_{1}=3}\\ \\ \boxed{x_{2}=-0.5}$

<h2>Explanation:</h2>

Hello! Recall that if you want to get correct and good answers, you need to post your question in a clear way. However, I'll try to help you, so let's get started:

The quadratic equation isn't clear, so let's assume that equation is as follows:

$2x^{2}-5x-3=0$

A quadratic equation is given by the form:

$ax^2+bx+c=0$

To find the solutions, we can apply the quadratic formula:

$x_{12}=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ \\ \\ Here: \\ \\ a=2 \\ b=-5 \\ c=-3 \\ \\ \\ Substituting: \\ \\ x_{12}=\frac{-(-5) \pm \sqrt{(-5)^2-4(2)(-3)}}{2(2)} \\ \\ x_{12}=\frac{5 \pm \sqrt{25+24}}{4} \\ \\ Finally, \ we \ get \ two \ solutions: \\ \\ x_{1}=\frac{5+ \sqrt{25+24}}{4}=\frac{12}{4} \therefore \boxed{x_{1}=3} \\ \\ \\ x_{2}=\frac{5-\sqrt{25+24}}{4}=-\frac{2}{4} \therefore \boxed{x_{2}=-0.5}$