# How do you solve 7(x^2-2x+3) = -3(x^2-5x)?

##### 2 Answers
Mar 27, 2015

Let's do the multiplications at both sides:
$7 {x}^{2} - 14 x + 21 = - 3 {x}^{2} + 15 x$
Add $3 {x}^{2}$ at both sides:
$10 {x}^{2} - 14 x + 21 = 15 x$
Subtract $15 x$ from both sides:
$10 {x}^{2} - 29 x + 21 = 0$

So we have a quadratic equation of the form $a {x}^{2} + b x + c$, where $a = 10$, $b = - 29$, and $c = 21$. Let's calculate the discriminant:
$\setminus \Delta = {b}^{2} - 4 a c = \left(- {29}^{2}\right) - 4 \cdot 10 \cdot 21 = 841 - 840 = 1$
Since $\setminus \Delta > 0$, there are two solutions ${x}_{1}$ and ${x}_{2}$, given by
${x}_{1 , 2} = \setminus \frac{- b \setminus \pm \setminus \sqrt{\setminus \Delta}}{2 a}$
Since $\setminus \Delta = 1$, its square root is also 1. The two solutions are thus
$\setminus \frac{29 \setminus \pm 1}{20}$
So, ${x}_{1} = \frac{29 + 1}{20} = \frac{3}{2}$ is the first solution, while ${x}_{2} = \frac{29 - 1}{20} = \frac{7}{5}$ is the second.

Mar 27, 2015

Have a look: