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3x^2-10x-48=0
a = 3; b = -10; c = -48;
Δ = b2-4ac
Δ = -102-4·3·(-48)
Δ = 676
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{676}=26$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-26}{2*3}=\frac{-16}{6} =-2+2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+26}{2*3}=\frac{36}{6} =6 $
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