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4x^2+x-14=0
a = 4; b = 1; c = -14;
Δ = b2-4ac
Δ = 12-4·4·(-14)
Δ = 225
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{225}=15$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-15}{2*4}=\frac{-16}{8} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+15}{2*4}=\frac{14}{8} =1+3/4 $
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