How do you solve #x^2+7x=-3# using the quadratic formula?

Question

Eli Assouline · Accepted Answer

Rewrite the given formula in standard form, then apply the quadratic formula for roots to get #x=-1/2# or #x=-13/2# #x^2+7x=-3# #color(white)("XXXX")#is equivalent to #x^2+7x+3=0# #color(white)("XXXX")#which is a quadratic in standard form #color(white)("XXXX")##color(white)("XXXX")##ax^2+bx+c=0# #color(white)("XXXX")#whose roots are given by the quadratic formula: #color(white)("XXXX")##color(white)("XXXX")##x = (-b+-sqrt(b^2-4ac))/(2a)# Using #a=1#, #b=7#, and #c=3# we have #color(white)("XXXX")##x= (-7+-sqrt(7^2-4(1)(3)))/(2(1)# #color(white)("XXXX")##x = (-7+-sqrt(36))/2# #color(white)("XXXX")##x = (-7+-6)/2# #x=-1/2# or #x=-13/2#

Kennedy Adams · Answer

undefined To solve the equation $x^2 + 7x = -3$ using the quadratic formula:

1. Identify the coefficients: $a = 1$, $b = 7$, and $c = -3$.
2. Apply the quadratic formula: $x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$.
3. Substitute the values of $a$, $b$, and $c$ into the formula.
4. Solve for $x$ by calculating the discriminant $b^2 - 4ac$.
5. Substitute the discriminant value into the formula and calculate the two possible values of $x$.