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

Question

Ethan Adkins · Accepted Answer

The solutions for the equation are:
#color(blue)(x=-1 , x=-2# #2x^2+6x+4=0# The equation is of the form #color(blue)(ax^2+bx+c=0# where: #a=2, b=6, c=4# The Discriminant is given by: #Delta=b^2-4*a*c# # = (6)^2-(4*(2)*4)# # = 36 -32 = 4# The solutions are found using the formula #x=(-b+-sqrtDelta)/(2*a)# #x = ((-6)+-sqrt(4))/(2*2) = ((-6+-2))/4# #x = ((-6+2))/4 = -4/4 = -1# #x = ((-6-2))/4 = -8/4 = -2# #color(blue)(x=-1 , x=-2#

Eva Abramson · Answer

undefined To solve the quadratic equation $2x^2 + 6x + 4 = 0$ using the quadratic formula, we first identify the coefficients: $a = 2$, $b = 6$, and $c = 4$. Then, we plug these values into the quadratic formula:

$$x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$$

Substitute the values of $a$, $b$, and $c$:

$$x = \frac{{-6 \pm \sqrt{{6^2 - 4 \cdot 2 \cdot 4}}}}{{2 \cdot 2}}$$

Simplify the expression under the square root:

$$x = \frac{{-6 \pm \sqrt{{36 - 32}}}}{{4}}$$

$$x = \frac{{-6 \pm \sqrt{{4}}}}{{4}}$$

$$x = \frac{{-6 \pm 2}}{{4}}$$

Now, we have two possible solutions:

$$x_1 = \frac{{-6 + 2}}{{4}} = \frac{{-4}}{{4}} = -1$$

$$x_2 = \frac{{-6 - 2}}{{4}} = \frac{{-8}}{{4}} = -2$$

Therefore, the solutions to the equation $2x^2 + 6x + 4 = 0$ are $x = -1$ and $x = -2$.