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4x^2+21x-4=0
a = 4; b = 21; c = -4;
Δ = b2-4ac
Δ = 212-4·4·(-4)
Δ = 505
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-\sqrt{505}}{2*4}=\frac{-21-\sqrt{505}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+\sqrt{505}}{2*4}=\frac{-21+\sqrt{505}}{8} $
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