urgent

3R+R^2=0.83 what is te way of solving

Formula:

\(\begin{array}{|rcll|} \hline ax^2+bx+c &=& 0 \\ x &=& {-b \pm \sqrt{b^2-4ac} \over 2a} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline 3R+R^2 &=& 0.83 & | \quad -0.83 \\ R^2 + 3R -0.83 &=& 0 & | \qquad a = 1 \qquad b = 3 \qquad c = -0.83 \\ R &=& \frac{-3 \pm \sqrt{3^2-4\cdot 1\cdot( -0.83 )} } { 2\cdot 1} \\ R &=& \frac{-3 \pm \sqrt{9+4\cdot 0.83 } } { 2 } \\ R &=& \frac{-3 \pm \sqrt{9+3.32 } } { 2 } \\ R &=& \frac{-3 \pm \sqrt{12.32 } } { 2 } \\ R &=& \frac{-3 \pm 3.50998575496 } { 2 } \\\\ R_1 &=& \frac{-3 + 3.50998575496 } { 2 } \\ R_1 &=& \frac{0.50998575496 } { 2 } \\ \mathbf{R_1} & \mathbf{=} & \mathbf{0.25499287748} \\\\ R_2 &=& \frac{-3 - 3.50998575496 } { 2 } \\ R_2 &=& \frac{-6.50998575496 } { 2 } \\ \mathbf{R_2} & \mathbf{=} & \mathbf{-3.25499287748} \\ \hline \end{array}\)

Re-arrange to    r^2 +3r - 0.83 = 0

Now use the Quadratic Formula

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

(-3 +-sqrt(9+3.32)  )  / 2 = -3/2 +-sqrt(12.32)/2   =- 3/2+- 1.1916  =   .254     ,  3.255

(corrected)