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21x^2+43x-14=0
a = 21; b = 43; c = -14;
Δ = b2-4ac
Δ = 432-4·21·(-14)
Δ = 3025
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{3025}=55$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(43)-55}{2*21}=\frac{-98}{42} =-2+1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(43)+55}{2*21}=\frac{12}{42} =2/7 $
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