# How to solve 10x^2+25x -15=0 ?

## Question:

How to solve {eq}10x^2 +25x-15 = 0 {/eq} ?

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

{eq}\eqalign{ & {\text{We have the quadratic equation }}10{x^2} + 25x - 15 = 0.{\text{ }} \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 = 10,\,\,\,b = 25,\,{\text{ and }}\,c = - 15.{\text{ Thus:}} \cr & \,\,\,\,\,x = \frac{{ - 25 \pm \sqrt {{{\left( {25} \right)}^2} - 4 \cdot \left( {10} \right) \cdot \left( { - 15} \right)} }}{{2 \cdot 10}} \cr & {\text{Simplifying:}} \cr & \,\,\,\,\,x = \frac{{ - 25 \pm \sqrt {1225} }}{{20}} \cr & \,\,\,\,\,x = \frac{{ - 25 + 35}}{{20}},\,\,\,\,\,\,\,x = \frac{{ - 25 - 35}}{{20}} \cr & {\text{Thus:}} \cr & \,\,\,\,\,x = \frac{{10}}{{20}} = \frac{1}{2},\,\,\,\,\,\,\,x = \frac{{ - 60}}{{20}} = - 3 \cr & {\text{Therefore}}{\text{, the solutions for this equation are: }} \cr & \,\,\,\,\boxed{x = \frac{1}{2},\,\,\,\,x = - 3} \cr} {/eq}