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2x^2+4=x^2-5x
We move all terms to the left:
2x^2+4-(x^2-5x)=0
We get rid of parentheses
2x^2-x^2+5x+4=0
We add all the numbers together, and all the variables
x^2+5x+4=0
a = 1; b = 5; c = +4;
Δ = b2-4ac
Δ = 52-4·1·4
Δ = 9
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{9}=3$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-3}{2*1}=\frac{-8}{2} =-4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+3}{2*1}=\frac{-2}{2} =-1 $
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