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3x^2+4x-11=13=10x
We move all terms to the left:
3x^2+4x-11-(13)=0
We add all the numbers together, and all the variables
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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