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2x^2-4x-48=0
a = 2; b = -4; c = -48;
Δ = b2-4ac
Δ = -42-4·2·(-48)
Δ = 400
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{400}=20$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-20}{2*2}=\frac{-16}{4} =-4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+20}{2*2}=\frac{24}{4} =6 $
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