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3x^2+2x=x(5x-5)
We move all terms to the left:
3x^2+2x-(x(5x-5))=0
We calculate terms in parentheses: -(x(5x-5)), so:We get rid of parentheses
x(5x-5)
We multiply parentheses
5x^2-5x
Back to the equation:
-(5x^2-5x)
3x^2-5x^2+2x+5x=0
We add all the numbers together, and all the variables
-2x^2+7x=0
a = -2; b = 7; c = 0;
Δ = b2-4ac
Δ = 72-4·(-2)·0
Δ = 49
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{49}=7$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-7}{2*-2}=\frac{-14}{-4} =3+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+7}{2*-2}=\frac{0}{-4} =0 $
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