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x^2+8x-5=9x-3
We move all terms to the left:
x^2+8x-5-(9x-3)=0
We get rid of parentheses
x^2+8x-9x+3-5=0
We add all the numbers together, and all the variables
x^2-1x-2=0
a = 1; b = -1; c = -2;
Δ = b2-4ac
Δ = -12-4·1·(-2)
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-3}{2*1}=\frac{-2}{2} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+3}{2*1}=\frac{4}{2} =2 $
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