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10x^2+40x-77=0
a = 10; b = 40; c = -77;
Δ = b2-4ac
Δ = 402-4·10·(-77)
Δ = 4680
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4680}=\sqrt{36*130}=\sqrt{36}*\sqrt{130}=6\sqrt{130}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-6\sqrt{130}}{2*10}=\frac{-40-6\sqrt{130}}{20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+6\sqrt{130}}{2*10}=\frac{-40+6\sqrt{130}}{20} $
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