Answer:
Quadratic Equation:
[tex]3x^2=2x +5[/tex]
[tex]\text{Standard Form: } 3x^2-2x-5=0[/tex]
From the standard form of a Quadratic Function, we get:
[tex]a=3,\:b=-2,\:c=-5[/tex]
Discriminant:
[tex]\Delta=\left(-2\right)^2-4\cdot \:3\left(-5\right)[/tex]
[tex]\Delta=\left(-2\right)^2+4\cdot \:3\cdot \:5[/tex]
[tex]\Delta=64[/tex]
From the discriminant, we conclude that the equation will have two real solutions.
State that:
[tex]b^2-4ac <0:\text{The equation has no real solutions, the solutions are complex}[/tex]
[tex]b^2-4ac =0:\text{The equation has 1 real solution}[/tex]
[tex]b^2-4ac >0:\text{The equation has 2 real solutions}[/tex]
By the way, solving the equation given:
[tex]$x=\frac{2\pm\sqrt{64}}{2\cdot \:3}$[/tex]
[tex]$x=\frac{2\pm\sqrt{64}}{6}$[/tex]
[tex]$x=\frac{2\pm8}{6}$[/tex]
[tex]$x_{1} =\frac{10}{6}=\frac{5}{3} $[/tex]
[tex]$x_{2}=\frac{-6}{6} =-1$[/tex]
