15=2x^2+3x+10

Solution for 15=2x^2+3x+10 equation:

15=2x^2+3x+10

We move all terms to the left:

15-(2x^2+3x+10)=0

We get rid of parentheses

-2x^2-3x-10+15=0

We add all the numbers together, and all the variables

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

a = -2; b = -3; c = +5;
Δ = b2-4ac
Δ = -32-4·(-2)·5
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-7}{2*-2}=\frac{-4}{-4}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+7}{2*-2}=\frac{10}{-4}
=-2+1/2
$