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5x+68=x^2-11x+28
We move all terms to the left:
5x+68-(x^2-11x+28)=0
We get rid of parentheses
-x^2+5x+11x-28+68=0
We add all the numbers together, and all the variables
-1x^2+16x+40=0
a = -1; b = 16; c = +40;
Δ = b2-4ac
Δ = 162-4·(-1)·40
Δ = 416
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{416}=\sqrt{16*26}=\sqrt{16}*\sqrt{26}=4\sqrt{26}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{26}}{2*-1}=\frac{-16-4\sqrt{26}}{-2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{26}}{2*-1}=\frac{-16+4\sqrt{26}}{-2} $
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