24x^2=3x+27

Solution for 24x^2=3x+27 equation:

24x^2=3x+27

We move all terms to the left:

24x^2-(3x+27)=0

We get rid of parentheses

24x^2-3x-27=0

a = 24; b = -3; c = -27;
Δ = b2-4ac
Δ = -32-4·24·(-27)
Δ = 2601
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2601}=51$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-51}{2*24}=\frac{-48}{48}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+51}{2*24}=\frac{54}{48}
=1+1/8
$