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24=3x^2-4x
We move all terms to the left:
24-(3x^2-4x)=0
We get rid of parentheses
-3x^2+4x+24=0
a = -3; b = 4; c = +24;
Δ = b2-4ac
Δ = 42-4·(-3)·24
Δ = 304
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{304}=\sqrt{16*19}=\sqrt{16}*\sqrt{19}=4\sqrt{19}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{19}}{2*-3}=\frac{-4-4\sqrt{19}}{-6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{19}}{2*-3}=\frac{-4+4\sqrt{19}}{-6} $
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