+1439 Find the largest value of $x$ such that $3x^2 + 17x + 15 = 2x^2 + 21x + 12 - 5x^2 + 17x + 34.$
+1926 Combining all like terms and moving all terms to one side we get
\(6x^2-21x-31=0\)
Now, we simply use the quadratic equation, we get
\(x_{1,\:2}=\frac{-\left(-21\right)\pm \sqrt{\left(-21\right)^2-4\cdot \:6\left(-31\right)}}{2\cdot \:6}\)
\(x_1=\frac{-\left(-21\right)+\sqrt{1185}}{2\cdot \:6},\:x_2=\frac{-\left(-21\right)-\sqrt{1185}}{2\cdot \:6}\)
\(x=\frac{21+\sqrt{1185}}{12},\:x=\frac{21-\sqrt{1185}}{12}\)
So our final answers are \(x=\frac{21+\sqrt{1185}}{12},\:x=\frac{21-\sqrt{1185}}{12}\)
Thanks! :)
