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3x^2-3x=2x^2+12x
We move all terms to the left:
3x^2-3x-(2x^2+12x)=0
We get rid of parentheses
3x^2-2x^2-3x-12x=0
We add all the numbers together, and all the variables
x^2-15x=0
a = 1; b = -15; c = 0;
Δ = b2-4ac
Δ = -152-4·1·0
Δ = 225
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{225}=15$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-15}{2*1}=\frac{0}{2} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+15}{2*1}=\frac{30}{2} =15 $
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