# How do you use the quadratic formula to solve 3/4(x-1)^2=-4/3x+4/5?

Jun 20, 2017

Before one uses the quadratic formula, one must write the equation in the form:

$a {x}^{2} + b x + c = 0$

Then substitute, a, b, and c into the formula:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 \left(a\right) \left(c\right)}}{2 a}$

#### Explanation:

Given: $\frac{3}{4} {\left(x - 1\right)}^{2} = - \frac{4}{3} x + \frac{4}{5}$

Multiply both sides of the equation by the factors: $\left(3\right) \left(4\right) \left(5\right)$

$45 {\left(x - 1\right)}^{2} = - 80 x + 48$

Expand the square:

$45 \left({x}^{2} - 2 x + 1\right) = - 80 x + 48$

Distribute the 45:

$45 {x}^{2} - 90 x + 45 = - 80 x + 48$

Combine like terms:

$45 {x}^{2} - 10 x - 3 = 0$

By observation, $a = 45 , b = - 10 , \mathmr{and} c = - 3$

Substitute into the formula:

$x = \frac{10 \pm \sqrt{{\left(- 10\right)}^{2} - 4 \left(45\right) \left(- 3\right)}}{2 \left(45\right)}$

$x = \frac{10 \pm \sqrt{640}}{90}$

$x = \frac{10 \pm 8 \sqrt{10}}{90}$

$x = \frac{10 - 8 \sqrt{10}}{90}$ and $x = \frac{10 + 8 \sqrt{10}}{90}$