How do you solve #2x²+3x=5 # using the quadratic formula?

Question

Nathan Axel · Accepted Answer

#- 5/2, 1#

Bring the equation to standard form:

#y = 2x^2 + 3x - 5 = 0#.

Use the improved quadratic formula (Socratic Search)(https://tutor.hix.ai):

#D = d^2 = b^2 - 4ac = 9 + 40 = 49# --> #d = +- 7#

There are 2 real roots:

#x = - b/(2a) +- d/(2a) = - 3/4 +- 7 /4#

#x1 =( - 3 - 7)/4 = - 10/4 = -5/2#

#x2 = (- 3 + 7)/4 = 4/4 = 1#

Eva Adamson · Answer

undefined To solve the equation $2x^2 + 3x = 5$ using the quadratic formula, you first need to rearrange the equation into the standard form $ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are coefficients. In this case, $a = 2$, $b = 3$, and $c = -5$. Then, apply the quadratic formula:

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

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

$$x = \frac{{-3 \pm \sqrt{{3^2 - 4 \cdot 2 \cdot (-5)}}}}{{2 \cdot 2}}$$

Calculate the discriminant:

$$b^2 - 4ac = 3^2 - 4 \cdot 2 \cdot (-5) = 9 + 40 = 49$$

Plug the discriminant back into the quadratic formula:

$$x = \frac{{-3 \pm \sqrt{{49}}}}{{4}}$$

This gives two solutions:

$$x_1 = \frac{{-3 + 7}}{{4}} = \frac{4}{4} = 1$$

$$x_2 = \frac{{-3 - 7}}{{4}} = \frac{-10}{4} = -\frac{5}{2}$$

Therefore, the solutions to the equation are $x = 1$ and $x = -\frac{5}{2}$.