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3x^2+36x=42
We move all terms to the left:
3x^2+36x-(42)=0
a = 3; b = 36; c = -42;
Δ = b2-4ac
Δ = 362-4·3·(-42)
Δ = 1800
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1800}=\sqrt{900*2}=\sqrt{900}*\sqrt{2}=30\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-30\sqrt{2}}{2*3}=\frac{-36-30\sqrt{2}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+30\sqrt{2}}{2*3}=\frac{-36+30\sqrt{2}}{6} $
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