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15x=x^2+26
We move all terms to the left:
15x-(x^2+26)=0
We get rid of parentheses
-x^2+15x-26=0
We add all the numbers together, and all the variables
-1x^2+15x-26=0
a = -1; b = 15; c = -26;
Δ = b2-4ac
Δ = 152-4·(-1)·(-26)
Δ = 121
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{121}=11$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-11}{2*-1}=\frac{-26}{-2} =+13 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+11}{2*-1}=\frac{-4}{-2} =+2 $
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