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-2x^2+48=4x
We move all terms to the left:
-2x^2+48-(4x)=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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