# How do you use the quadratic formula to solve 2.5x^2-3.3x=1.2x-1.5?

Sep 10, 2017

Solution: $x \approx 1.36 , x \approx 0.44$

#### Explanation:

$2.5 {x}^{2} - 3.3 x = 1.2 x - 1.5 \mathmr{and} 2.5 {x}^{2} - 3.3 x - 1.2 x + 1.5 = 0$ or

$2.5 {x}^{2} - 4.5 x + 1.5 = 0$ Comparing with standard quadratic

equation $a {x}^{2} + b x + c = 0$ we get here,

$a = 2.5 , b = - 4.5 , c = 1.5$. Discriminant $D = {b}^{2} - 4 a c$ or

$D = - {4.5}^{2} - 4 \cdot 2.5 \cdot 1.5 = 5.25$

$x = \frac{- b \pm \sqrt{D}}{2 a} = \frac{4.5 \pm \sqrt{5.25}}{2 \cdot 2.5}$ or

$x = \frac{4.5 \pm \sqrt{5.25}}{5} \mathmr{and} x \approx 0.9 \pm \frac{2.29}{5}$ or

$x \approx 0.9 \pm \frac{2.29}{5} \mathmr{and} x \approx 0.9 \pm 0.46$ or

$x \approx 1.36 \left(2 \mathrm{dp}\right) , x \approx 0.44 \left(2 \mathrm{dp}\right)$ [Ans]