# How do you find the discriminant and how many solutions does 2x^2-7x-4=0 have?

Discriminant $= 81$ & equation has $2$ solutions i.e. two real distinct roots

#### Explanation:

The given quadratic equation: $2 {x}^{2} - 7 x - 4 = 0$ will always have two solutions .

Now, comparing with $a {x}^{2} + b x + c = 0$, we get

$a = 2 , b = - 7$ & $c = - 4$

The discriminant $\setminus \Delta = {b}^{2} - 4 a c$ is given as

${b}^{2} - 4 a c = {\left(- 7\right)}^{2} - 4 \left(2\right) \left(- 4\right) = 81 > 0$

The given quadratic equation has two real distinct roots.

Jul 9, 2018

Two Solutions: $x = 4 \mathmr{and} x = - 0.5$

#### Explanation:

$2 {x}^{2} - 7 x - 4 = 0$

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

$a = 2 , b = - 7 , c = - 4$ Discriminant $D = {b}^{2} - 4 a c$

$D = 49 + 32 = 81$, discriminant is positive, we get two real

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

$x = \frac{7 \pm \sqrt{81}}{4} = \frac{7 \pm 9}{4} \therefore x = \frac{7 + 9}{4} = \frac{16}{4} = 4$ and

$x = \frac{7 - 9}{4} = - \frac{2}{4} = - 0.5 \therefore x = 4 \mathmr{and} x = - 0.5$

Solution: $x = 4 \mathmr{and} x = - 0.5$ [Ans]