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c^2-15c+13=-6c
We move all terms to the left:
c^2-15c+13-(-6c)=0
We get rid of parentheses
c^2-15c+6c+13=0
We add all the numbers together, and all the variables
c^2-9c+13=0
a = 1; b = -9; c = +13;
Δ = b2-4ac
Δ = -92-4·1·13
Δ = 29
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-\sqrt{29}}{2*1}=\frac{9-\sqrt{29}}{2} $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{29}}{2*1}=\frac{9+\sqrt{29}}{2} $
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