Question

$5 PAYPAL + BRAINLIEST TO ANSWER THS

Answer 1

Answer:

Quadratic Equation:

$3x^2=2x +5$

$\text{Standard Form: } 3x^2-2x-5=0$

From the standard form of a Quadratic Function, we get:

$a=3,\:b=-2,\:c=-5$

Discriminant:

$\Delta=\left(-2\right)^2-4\cdot \:3\left(-5\right)$

$\Delta=\left(-2\right)^2+4\cdot \:3\cdot \:5$

$\Delta=64$

From the discriminant, we conclude that the equation will have two real solutions.

State that:

$b^2-4ac$

$b^2-4ac =0:\text{The equation has 1 real solution}$

$b^2-4ac >0:\text{The equation has 2 real solutions}$

By the way, solving the equation given:

$x=\frac{2\pm\sqrt{64}}{2\cdot \:3}$

$x=\frac{2\pm\sqrt{64}}{6}$

$x=\frac{2\pm8}{6}$

$x_{1} =\frac{10}{6}=\frac{5}{3}$

$x_{2}=\frac{-6}{6} =-1$