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20x+3x^2=135
We move all terms to the left:
20x+3x^2-(135)=0
a = 3; b = 20; c = -135;
Δ = b2-4ac
Δ = 202-4·3·(-135)
Δ = 2020
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2020}=\sqrt{4*505}=\sqrt{4}*\sqrt{505}=2\sqrt{505}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{505}}{2*3}=\frac{-20-2\sqrt{505}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{505}}{2*3}=\frac{-20+2\sqrt{505}}{6} $
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