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