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(2x-5)(3x-10)=180
We move all terms to the left:
(2x-5)(3x-10)-(180)=0
We multiply parentheses ..
(+6x^2-20x-15x+50)-180=0
We get rid of parentheses
6x^2-20x-15x+50-180=0
We add all the numbers together, and all the variables
6x^2-35x-130=0
a = 6; b = -35; c = -130;
Δ = b2-4ac
Δ = -352-4·6·(-130)
Δ = 4345
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-\sqrt{4345}}{2*6}=\frac{35-\sqrt{4345}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+\sqrt{4345}}{2*6}=\frac{35+\sqrt{4345}}{12} $
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