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2x+3=x^2-2x-
We move all terms to the left:
2x+3-(x^2-2x-)=0
We get rid of parentheses
-x^2+2x+2x+3+=0
We add all the numbers together, and all the variables
-1x^2+4x=0
a = -1; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·(-1)·0
Δ = 16
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{16}=4$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*-1}=\frac{-8}{-2} =+4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*-1}=\frac{0}{-2} =0 $
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