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4x^2+49x-16=0
a = 4; b = 49; c = -16;
Δ = b2-4ac
Δ = 492-4·4·(-16)
Δ = 2657
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{-(49)-\sqrt{2657}}{2*4}=\frac{-49-\sqrt{2657}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(49)+\sqrt{2657}}{2*4}=\frac{-49+\sqrt{2657}}{8} $
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