# How do you solve -0.12x^2 + 1.4x - 3.5 = 0.5 using the quadratic formula?

Jul 9, 2015

I found:
${x}_{1} = 5$
${x}_{2} = \frac{20}{3}$

#### Explanation:

Ok...I`ll try to "simplify" it a bit to avoid decimals...(I do not like them!).
I can write it as:
$- \frac{12}{100} {x}^{2} + \frac{14}{10} x - \frac{35}{10} = \frac{5}{10}$
getting rid of the denominators (after taking $100$ as common) I got:
$- 12 {x}^{2} + 140 x - 400 = 0$
that is in the form:
$a {x}^{2} + b x + c = 0$
With:
$a = - 12$
$b = 140$
$c = - 400$
I now use the Quadratic Formula:
${x}_{1 , 2} = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} = \frac{- 140 \pm \sqrt{{140}^{2} - 4 \left(- 12 \cdot - 400\right)}}{2 \cdot - 12} =$
$= \frac{- 140 \pm \sqrt{400}}{-} 24 = \frac{- 140 \pm 20}{-} 24 =$
You get two solutions:
${x}_{1} = \frac{- 140 + 20}{-} 24 = 5$
${x}_{2} = \frac{- 140 - 20}{-} 24 = \frac{160}{24} = \frac{20}{3}$