The root of the quadratic equation x = -3/4[tex]\pm[/tex] √-31/ 4.
What is a solution for a quadratic equation?
Suppose that we've a function y = f(x) such that f(x) is quadratic.
When y = 0, then the values of x for which f(x) = 0 is called solution of quadratic equation f(x) = 0
These solution gives values of x, and when we plot x and f(x), we'd see that the graph intersects the x-axis at its solution points.
Then its roots are given as
[tex]x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]
Given; 2x^2 + 3x + 5 = 0
We know that a=2, b=3, c=5
[tex]x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]
[tex]x = \dfrac{-3\pm \sqrt{3^2 - 4(2) ( 5)}}{2(2)}\\\\\\x = \dfrac{-3\pm \sqrt{9 - 40}}{4}\\\\\\x = \dfrac{-3\pm \sqrt{-31}}{4}\\\\\\x = \dfrac{-3}{4}\pm \dfrac {\sqrt{-31}}{4}[/tex]
Learn more about solutions of a quadratic equation here:
https://brainly.com/question/15582302
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