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2x(3x-10)=180
We move all terms to the left:
2x(3x-10)-(180)=0
We multiply parentheses
6x^2-20x-180=0
a = 6; b = -20; c = -180;
Δ = b2-4ac
Δ = -202-4·6·(-180)
Δ = 4720
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4720}=\sqrt{16*295}=\sqrt{16}*\sqrt{295}=4\sqrt{295}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{295}}{2*6}=\frac{20-4\sqrt{295}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{295}}{2*6}=\frac{20+4\sqrt{295}}{12} $
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