c^2+7c-3=0

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Solution for c^2+7c-3=0 equation: 



c^2+7c-3=0

a = 1; b = 7; c = -3;
Δ = b2-4ac
Δ = 72-4·1·(-3)
Δ = 61
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{-(7)-\sqrt{61}}{2*1}=\frac{-7-\sqrt{61}}{2}
$


$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{61}}{2*1}=\frac{-7+\sqrt{61}}{2}
$

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