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x2+5x=2x^2-4x-20-190
We move all terms to the left:
x2+5x-(2x^2-4x-20-190)=0
We add all the numbers together, and all the variables
x^2+5x-(2x^2-4x-20-190)=0
We get rid of parentheses
x^2-2x^2+5x+4x+20+190=0
We add all the numbers together, and all the variables
-1x^2+9x+210=0
a = -1; b = 9; c = +210;
Δ = b2-4ac
Δ = 92-4·(-1)·210
Δ = 921
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{-(9)-\sqrt{921}}{2*-1}=\frac{-9-\sqrt{921}}{-2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{921}}{2*-1}=\frac{-9+\sqrt{921}}{-2} $
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