3x^2+10=x^2+23x-20

Solution for 3x^2+10=x^2+23x-20 equation:

3x^2+10=x^2+23x-20

We move all terms to the left:

3x^2+10-(x^2+23x-20)=0

We get rid of parentheses

3x^2-x^2-23x+20+10=0

We add all the numbers together, and all the variables

2x^2-23x+30=0

a = 2; b = -23; c = +30;
Δ = b2-4ac
Δ = -232-4·2·30
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-17}{2*2}=\frac{6}{4}
=1+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+17}{2*2}=\frac{40}{4}
=10
$