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48x^2+56x+15=0
a = 48; b = 56; c = +15;
Δ = b2-4ac
Δ = 562-4·48·15
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{256}=16$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(56)-16}{2*48}=\frac{-72}{96} =-3/4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(56)+16}{2*48}=\frac{-40}{96} =-5/12 $
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