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x^2+20-9x=56
We move all terms to the left:
x^2+20-9x-(56)=0
We add all the numbers together, and all the variables
x^2-9x-36=0
a = 1; b = -9; c = -36;
Δ = b2-4ac
Δ = -92-4·1·(-36)
Δ = 225
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{225}=15$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-15}{2*1}=\frac{-6}{2} =-3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+15}{2*1}=\frac{24}{2} =12 $
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