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x^2-13x+24=-x+4
We move all terms to the left:
x^2-13x+24-(-x+4)=0
We add all the numbers together, and all the variables
x^2-13x-(-1x+4)+24=0
We get rid of parentheses
x^2-13x+1x-4+24=0
We add all the numbers together, and all the variables
x^2-12x+20=0
a = 1; b = -12; c = +20;
Δ = b2-4ac
Δ = -122-4·1·20
Δ = 64
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{64}=8$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-8}{2*1}=\frac{4}{2} =2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+8}{2*1}=\frac{20}{2} =10 $
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