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3x^2=11x+24
We move all terms to the left:
3x^2-(11x+24)=0
We get rid of parentheses
3x^2-11x-24=0
a = 3; b = -11; c = -24;
Δ = b2-4ac
Δ = -112-4·3·(-24)
Δ = 409
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-\sqrt{409}}{2*3}=\frac{11-\sqrt{409}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{409}}{2*3}=\frac{11+\sqrt{409}}{6} $
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