# How do you solve \frac { x - 2+ 3} { x + 3} + \frac { 3} { 20} = x?

Sep 22, 2017

Solution : $x \approx - 2.44 , x = 0.59$

#### Explanation:

$\frac{x - 2 + 3}{x + 3} + \frac{3}{20} = x$ or

 (x+1)/(x+3) +3/20 =x ; Multiplying by $20 \left(x + 3\right)$ on both sides,

we get $20 \left(x + 1\right) + 3 \left(x + 3\right) = 20 x \left(x + 3\right)$ or

$20 x + 20 + 3 x + 9 = 20 {x}^{2} + 60 x$ or

$20 {x}^{2} + 60 x - 20 x - 3 x - 29 = 0$ or

 20x^2+37x -29 =0 ; a=20 , b= 37 ;c =-29

Discriminant : $D = {b}^{2} - 4 a c = {37}^{2} - 4 \cdot 20 \cdot \left(- 29\right) = 3689$

$x = \frac{- b \pm \sqrt{D}}{2 a} = \frac{- 37 \pm \sqrt{3689}}{2 \cdot 20}$ or

$x = - \frac{37}{40} \pm \frac{60.74}{40} \therefore x \approx - 2.44 , x = 0.59$

Solution : $x \approx - 2.44 , x = 0.59$ [Ans]