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7x^2+6x-3=0
a = 7; b = 6; c = -3;
Δ = b2-4ac
Δ = 62-4·7·(-3)
Δ = 120
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{120}=\sqrt{4*30}=\sqrt{4}*\sqrt{30}=2\sqrt{30}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{30}}{2*7}=\frac{-6-2\sqrt{30}}{14} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{30}}{2*7}=\frac{-6+2\sqrt{30}}{14} $
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