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216=4x+3x^2
We move all terms to the left:
216-(4x+3x^2)=0
We get rid of parentheses
-3x^2-4x+216=0
a = -3; b = -4; c = +216;
Δ = b2-4ac
Δ = -42-4·(-3)·216
Δ = 2608
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2608}=\sqrt{16*163}=\sqrt{16}*\sqrt{163}=4\sqrt{163}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{163}}{2*-3}=\frac{4-4\sqrt{163}}{-6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{163}}{2*-3}=\frac{4+4\sqrt{163}}{-6} $
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