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x2+20x=224
We move all terms to the left:
x2+20x-(224)=0
We add all the numbers together, and all the variables
x^2+20x-224=0
a = 1; b = 20; c = -224;
Δ = b2-4ac
Δ = 202-4·1·(-224)
Δ = 1296
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{1296}=36$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-36}{2*1}=\frac{-56}{2} =-28 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+36}{2*1}=\frac{16}{2} =8 $
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