X^2-15x+24=3x+4

Solution for X^2-15x+24=3x+4 equation:

X^2-15X+24=3X+4

We move all terms to the left:

X^2-15X+24-(3X+4)=0

We get rid of parentheses

X^2-15X-3X-4+24=0

We add all the numbers together, and all the variables

X^2-18X+20=0

a = 1; b = -18; c = +20;
Δ = b2-4ac
Δ = -182-4·1·20
Δ = 244
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:

$\sqrt{\Delta}=\sqrt{244}=\sqrt{4*61}=\sqrt{4}*\sqrt{61}=2\sqrt{61}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{61}}{2*1}=\frac{18-2\sqrt{61}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{61}}{2*1}=\frac{18+2\sqrt{61}}{2}
$