3x+5=2x^2-3x

Solution for 3x+5=2x^2-3x equation:

3x+5=2x^2-3x

We move all terms to the left:

3x+5-(2x^2-3x)=0

We get rid of parentheses

-2x^2+3x+3x+5=0

We add all the numbers together, and all the variables

-2x^2+6x+5=0

a = -2; b = 6; c = +5;
Δ = b2-4ac
Δ = 62-4·(-2)·5
Δ = 76
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{76}=\sqrt{4*19}=\sqrt{4}*\sqrt{19}=2\sqrt{19}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{19}}{2*-2}=\frac{-6-2\sqrt{19}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{19}}{2*-2}=\frac{-6+2\sqrt{19}}{-4}
$