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1120=x(x-3)
We move all terms to the left:
1120-(x(x-3))=0
We calculate terms in parentheses: -(x(x-3)), so:We get rid of parentheses
x(x-3)
We multiply parentheses
x^2-3x
Back to the equation:
-(x^2-3x)
-x^2+3x+1120=0
We add all the numbers together, and all the variables
-1x^2+3x+1120=0
a = -1; b = 3; c = +1120;
Δ = b2-4ac
Δ = 32-4·(-1)·1120
Δ = 4489
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{4489}=67$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-67}{2*-1}=\frac{-70}{-2} =+35 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+67}{2*-1}=\frac{64}{-2} =-32 $
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