# How do you solve 5x^2-6x-4=0 using the quadratic formula?

Apr 13, 2018

#### Answer:

The roots are $\frac{3 + \sqrt{29}}{5}$ or $\frac{3 - \sqrt{29}}{5}$.

#### Explanation:

x=(-b±sqrt(b^2-4ac))/(2a)

x=6±sqrt((-6)^2-4*5*(-4))/(2*5)

x=(6±sqrt(36+80))/(10)

x=(6±sqrt(116))/(10)

x=(6±2sqrt(29))/(10)

x=(3±sqrt(29))/(5)

Therefore the roots can be $\frac{3 + \sqrt{29}}{5}$ or $\frac{3 - \sqrt{29}}{5}$.

Apr 13, 2018

#### Answer:

Solution: $x = \frac{3}{5} \pm \frac{\sqrt{29}}{5}$

#### Explanation:

$5 {x}^{2} - 6 x - 4 = 0$

Comparing with standard quadratic equation $a {x}^{2} + b x + c = 0$

$a = 5 , b = - 6 , c = - 4$. Discriminant, $D = {b}^{2} - 4 a c$ or

$D = 36 + 80 = 116$, discriminant is positive, we get two real

solutions, Quadratic formula: $x = \frac{- b \pm \sqrt{D}}{2 a}$or

$x = \frac{6 \pm \sqrt{116}}{10} = \frac{3}{5} \pm \frac{\sqrt{29}}{5}$

Solution: $x = \frac{3}{5} \pm \frac{\sqrt{29}}{5}$ [Ans]