# How do you use the quadratic formula to solve 1/2x^2-3x+2=3x-1?

Mar 28, 2017

${x}_{1} \approx 11.477$
${x}_{2} \approx 0.523$

#### Explanation:

Making right side of the equation equate as 0, we end up with:
$\frac{1}{2} {x}^{2} - 6 x + 3 = 0$

We can make it easier to visualise by multiplying both sides by 2,
${x}^{2} - 12 x + 6 = 0$

How many and what kinds of roots will we end up with?
${b}^{2} - 4 a c = {\left(- 12\right)}^{2} - 4 \left(1\right) \left(6\right) = 120 > 0$ (i.e. 2 real roots)

Let's find out what they are:

${x}_{1 , 2} = - \frac{b}{2 a} \pm \frac{\sqrt{{b}^{2} - 4 a c}}{2 a}$

${x}_{1} = - \frac{- 12}{2 \cdot 1} + \frac{\sqrt{120}}{2 \cdot 1} \approx 11.477$
${x}_{2} = - \frac{- 12}{2 \cdot 1} - \frac{\sqrt{120}}{2 \cdot 1} \approx 0.523$