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(2x-1)2=x2+3x-5
We move all terms to the left:
(2x-1)2-(x2+3x-5)=0
We add all the numbers together, and all the variables
-(+x^2+3x-5)+(2x-1)2=0
We multiply parentheses
-(+x^2+3x-5)+4x-2=0
We get rid of parentheses
-x^2-3x+4x+5-2=0
We add all the numbers together, and all the variables
-1x^2+x+3=0
a = -1; b = 1; c = +3;
Δ = b2-4ac
Δ = 12-4·(-1)·3
Δ = 13
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{13}}{2*-1}=\frac{-1-\sqrt{13}}{-2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{13}}{2*-1}=\frac{-1+\sqrt{13}}{-2} $
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