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15-4x=x^2-6x
We move all terms to the left:
15-4x-(x^2-6x)=0
We get rid of parentheses
-x^2-4x+6x+15=0
We add all the numbers together, and all the variables
-1x^2+2x+15=0
a = -1; b = 2; c = +15;
Δ = b2-4ac
Δ = 22-4·(-1)·15
Δ = 64
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{64}=8$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-8}{2*-1}=\frac{-10}{-2} =+5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+8}{2*-1}=\frac{6}{-2} =-3 $
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