Question

Solve the equation using quadratic formula.

$4r^2+3r-5=0$

Short answer

The roots are$r=\frac{-3+\sqrt{89}}{8},r=\frac{-3-\sqrt{89}}{8}$.

Step-by-step solution

 Given Information.    

The given equation is$4r^2+3r-5=0$.

Write the quadratic equation in standard form and identify a,b,c.

$4r^2+3r-5=0$

The equation is in standard form.

So,

$a=4,b=3,c=-5$

Substitute the values in quadratic formula.   

$$\begin{gathered} r=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ r=\frac{-3\pm \sqrt{3^2-4\left(4\right)(-5)}}{2\left(4\right)} \\ r=\frac{-3\pm \sqrt{9+80}}{8} \\ r=\frac{-3\pm \sqrt{89}}{8} \end{gathered}$$

Rewrite to show two solutions.

$$\begin{gathered} r=\frac{-3+\sqrt{89}}{8}, \\ r=\frac{-3-\sqrt{89}}{8} \end{gathered}$$