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x2-13x+8=7x
We move all terms to the left:
x2-13x+8-(7x)=0
We add all the numbers together, and all the variables
x^2-20x+8=0
a = 1; b = -20; c = +8;
Δ = b2-4ac
Δ = -202-4·1·8
Δ = 368
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{368}=\sqrt{16*23}=\sqrt{16}*\sqrt{23}=4\sqrt{23}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{23}}{2*1}=\frac{20-4\sqrt{23}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{23}}{2*1}=\frac{20+4\sqrt{23}}{2} $
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