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11x+33=x^2+3x
We move all terms to the left:
11x+33-(x^2+3x)=0
We get rid of parentheses
-x^2+11x-3x+33=0
We add all the numbers together, and all the variables
-1x^2+8x+33=0
a = -1; b = 8; c = +33;
Δ = b2-4ac
Δ = 82-4·(-1)·33
Δ = 196
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{196}=14$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-14}{2*-1}=\frac{-22}{-2} =+11 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+14}{2*-1}=\frac{6}{-2} =-3 $
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