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7x^2-3x-4=0
a = 7; b = -3; c = -4;
Δ = b2-4ac
Δ = -32-4·7·(-4)
Δ = 121
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{121}=11$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-11}{2*7}=\frac{-8}{14} =-4/7 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+11}{2*7}=\frac{14}{14} =1 $
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