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5b^2+19b+12=0
a = 5; b = 19; c = +12;
Δ = b2-4ac
Δ = 192-4·5·12
Δ = 121
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{121}=11$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-11}{2*5}=\frac{-30}{10} =-3 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+11}{2*5}=\frac{-8}{10} =-4/5 $
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