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50+2x(3x-10)=180
We move all terms to the left:
50+2x(3x-10)-(180)=0
We add all the numbers together, and all the variables
2x(3x-10)-130=0
We multiply parentheses
6x^2-20x-130=0
a = 6; b = -20; c = -130;
Δ = b2-4ac
Δ = -202-4·6·(-130)
Δ = 3520
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3520}=\sqrt{64*55}=\sqrt{64}*\sqrt{55}=8\sqrt{55}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-8\sqrt{55}}{2*6}=\frac{20-8\sqrt{55}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+8\sqrt{55}}{2*6}=\frac{20+8\sqrt{55}}{12} $
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