Expand.
\[2{x}^{2}-5x+20x-50=4x-15deg\]
Simplify \(2{x}^{2}-5x+20x-50\) to \(2{x}^{2}+15x-50\).
\[2{x}^{2}+15x-50=4x-15deg\]
Move all terms to one side.
\[2{x}^{2}+15x-50-4x+15deg=0\]
Simplify \(2{x}^{2}+15x-50-4x+15deg\) to \(2{x}^{2}+11x-50+15deg\).
\[2{x}^{2}+11x-50+15deg=0\]
Use the Quadratic Formula.
In general, given \(a{x}^{2}+bx+c=0\), there exists two solutions where:
\[x=\frac{-b+\sqrt{{b}^{2}-4ac}}{2a},\frac{-b-\sqrt{{b}^{2}-4ac}}{2a}\]
In this case, \(a=2\), \(b=11\) and \(c=-50+15deg\).
\[{x}^{}=\frac{-11+\sqrt{{11}^{2}-4\times 2(-50+15deg)}}{2\times 2},\frac{-11-\sqrt{{11}^{2}-4\times 2(-50+15deg)}}{2\times 2}\]
Simplify.
\[x=\frac{-11+\sqrt{521-120deg}}{4},\frac{-11-\sqrt{521-120deg}}{4}\]
\[x=\frac{-11+\sqrt{521-120deg}}{4},\frac{-11-\sqrt{521-120deg}}{4}\]
x=(-11+sqrt(521-120*deg))/4,(-11-sqrt(521-120*deg))/4
