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4b^2-13b-12=0
a = 4; b = -13; c = -12;
Δ = b2-4ac
Δ = -132-4·4·(-12)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{361}=19$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-19}{2*4}=\frac{-6}{8} =-3/4 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+19}{2*4}=\frac{32}{8} =4 $
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