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4k^2+19k+15=0
a = 4; b = 19; c = +15;
Δ = b2-4ac
Δ = 192-4·4·15
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{121}=11$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-11}{2*4}=\frac{-30}{8} =-3+3/4 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+11}{2*4}=\frac{-8}{8} =-1 $
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