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3x^2-7x+36=2x^2+6x
We move all terms to the left:
3x^2-7x+36-(2x^2+6x)=0
We get rid of parentheses
3x^2-2x^2-7x-6x+36=0
We add all the numbers together, and all the variables
x^2-13x+36=0
a = 1; b = -13; c = +36;
Δ = b2-4ac
Δ = -132-4·1·36
Δ = 25
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{25}=5$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-5}{2*1}=\frac{8}{2} =4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+5}{2*1}=\frac{18}{2} =9 $
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