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4a^2-3a-7=0
a = 4; b = -3; c = -7;
Δ = b2-4ac
Δ = -32-4·4·(-7)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{121}=11$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-11}{2*4}=\frac{-8}{8} =-1 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+11}{2*4}=\frac{14}{8} =1+3/4 $
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