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x^2+(5x)-(33)=4x+9
We move all terms to the left:
x^2+(5x)-(33)-(4x+9)=0
We get rid of parentheses
x^2+5x-4x-9-33=0
We add all the numbers together, and all the variables
x^2+x-42=0
a = 1; b = 1; c = -42;
Δ = b2-4ac
Δ = 12-4·1·(-42)
Δ = 169
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{169}=13$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-13}{2*1}=\frac{-14}{2} =-7 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+13}{2*1}=\frac{12}{2} =6 $
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