Math Problem Solver

Solve by quadratic formula:

Subtract the right hand side from the left hand side of: $$150 = - \frac{44 x^{2}}{100} + x \left(6 \cdot \frac{1}{100} + 13\right) + 47 \cdot \frac{1}{100} + 54$$ The result is a quadratic equation: $$44 \cdot \frac{1}{100} x^{2} - x \left(6 \cdot \frac{1}{100} + 13\right) + \frac{9553}{100} = 0$$

Simplify the left hand side of the equation into standard form: $$\begin{aligned}\frac{44 x^{2}}{100} + \frac{9553}{100} - x \left(13 + 6 \cdot \frac{1}{100}\right) &= 0\\\frac{11 x^{2}}{25} + \frac{9553}{100} - x \left(13 + \frac{3}{50}\right) &= 0\\\frac{11 x^{2}}{25} + \frac{9553}{100} - \frac{x 653}{50} &= 0\\\frac{11 x^{2}}{25} + \frac{\left(-653\right) x}{50} + \frac{9553}{100} &= 0\\\frac{11 x^{2}}{25} - \frac{653 x}{50} + \frac{9553}{100} &= 0\end{aligned}$$

Given a quadratic equation $a x^{2} + b x + c = 0$, where $a$, $b$, $c$ are constants and $a \ne 0$, the solutions are given by the quadratic formula: $$x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$$ In this case $a=\frac{11}{25}$, $b=- \frac{653}{50}$, and $c=\frac{9553}{100}$.

The discriminant is the quantity under the square root sign in the quadratic formula, and its sign determines the number of solutions to the quadratic equation when the coefficients are real. The discriminant is:$$b^{2}-4ac = \left(- \frac{653}{50}\right)^{2} - \frac{44}{25} \cdot \frac{9553}{100}=\frac{6077}{2500} > 0$$ The discriminant is greater than zero, so this quadratic equation has two real solutions.

The two solutions are: $$x = \frac{\left(-1\right) \left(- \frac{653}{50}\right) + \sqrt{\frac{6077}{2500}}}{2 \cdot \frac{11}{25}} = \frac{653}{44} + \frac{\sqrt{6077}}{44} \approx 16.612616$$ $$x = \frac{\left(-1\right) \left(- \frac{653}{50}\right) - \sqrt{\frac{6077}{2500}}}{2 \cdot \frac{11}{25}} = \frac{653}{44} - \frac{\sqrt{6077}}{44} \approx 13.069202$$