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c^2+15c-34=0
a = 1; b = 15; c = -34;
Δ = b2-4ac
Δ = 152-4·1·(-34)
Δ = 361
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}$$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{361}=19$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-19}{2*1}=\frac{-34}{2} =-17 $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+19}{2*1}=\frac{4}{2} =2 $
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