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4a^2-19a-100=0
a = 4; b = -19; c = -100;
Δ = b2-4ac
Δ = -192-4·4·(-100)
Δ = 1961
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}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-\sqrt{1961}}{2*4}=\frac{19-\sqrt{1961}}{8} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{1961}}{2*4}=\frac{19+\sqrt{1961}}{8} $
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