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7w^2+11w-4=0
a = 7; b = 11; c = -4;
Δ = b2-4ac
Δ = 112-4·7·(-4)
Δ = 233
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-\sqrt{233}}{2*7}=\frac{-11-\sqrt{233}}{14} $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{233}}{2*7}=\frac{-11+\sqrt{233}}{14} $
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