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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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-56)-16}{2*48}=\frac{40}{96} =5/12 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-56)+16}{2*48}=\frac{72}{96} =3/4 $
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