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2x-3+3x^2=90
We move all terms to the left:
2x-3+3x^2-(90)=0
We add all the numbers together, and all the variables
3x^2+2x-93=0
a = 3; b = 2; c = -93;
Δ = b2-4ac
Δ = 22-4·3·(-93)
Δ = 1120
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1120}=\sqrt{16*70}=\sqrt{16}*\sqrt{70}=4\sqrt{70}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4\sqrt{70}}{2*3}=\frac{-2-4\sqrt{70}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4\sqrt{70}}{2*3}=\frac{-2+4\sqrt{70}}{6} $
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