# Solve the quadratic equation 98 + 10x - x^2= 0.

## Question:

Solve the quadratic equation {eq}98 + 10x - x^2= 0. {/eq}

A quadratic equation is easily recognizable, since it is a second-degree equation of the polynomial type, and it has the following simplified form: {eq}a{x^2} + bx + c = 0 {/eq}. The solutions to this type of equation are formed by the values of the variable {eq}x {/eq}, for which the quadratic expression is equal to 0 and is written as: {eq}x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} {/eq}.

{eq}\eqalign{ & {\text{We have the quadratic function }}\,98 + 10x - {x^2} = 0.{\text{ Rewriting }} \cr & {\text{the equation, we have }}\,{x^2} - 10x - 98 = 0. \cr & {\text{So}}{\text{, for a quadratic equation of the form }}a{x^2} + bx + c = 0\,{\text{ }} \cr & {\text{the solutions (roots) are given by the quadratic formula:}} \cr & \,\,\,\,\,x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \cr & {\text{In this particular case }}\,a = 1,\,\,\,b = - 10,\,{\text{ and }}\,c = - 98.{\text{ Thus:}} \cr & \,\,\,\,\,x = \frac{{ - \left( { - 10} \right) \pm \sqrt {{{\left( { - 10} \right)}^2} - 4 \cdot 1 \cdot \left( { - 98} \right)} }}{{2 \cdot 1}} \cr & {\text{Simplifying:}} \cr & \,\,\,\,\,x = \frac{{10 \pm \sqrt {492} }}{2} \cr & \,\,\,\,\,x = \frac{{10 \pm 2\sqrt {123} }}{2} \cr & \,\,\,\,\,x = 5 \pm \sqrt {123} \cr & {\text{Thus:}} \cr & \,\,\,\,\, \Rightarrow x = 5 + \sqrt {123} = 16.09,\,\,\,\,\,\,\,x = 5 - \sqrt {123} = - 6.09 \cr & {\text{Therefore}}{\text{, the solutions for this equation are: }}\,\boxed{x = 16.09,\,\,\,\,x = - 6.09} \cr} {/eq}