How do you solve the equation by factoring or the quadratic formula #x^2+ 3x - 28 = 0#?

Question

Eva Adamson · Accepted Answer

Find the factors of 28 that will subtract to give 3 i.e. 4 and 7. Then factorise the expression

#x^2+3x-28 = (x+4)(x-7)# Therefore #x=-4# or #x=7#

Isaiah Anthony · Answer

undefined To solve the quadratic equation $x^2 + 3x - 28 = 0$, we can use either factoring or the quadratic formula.

1. Factoring:
   We need to find two numbers that multiply to -28 and add up to 3.
   The numbers are 7 and -4.
   So, we rewrite the equation as $(x + 7)(x - 4) = 0$.
   Setting each factor equal to zero, we get $x + 7 = 0$ and $x - 4 = 0$.
   Solving these equations gives us $x = -7$ and $x = 4$.

2. Quadratic formula:
   The quadratic formula is $x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$, where $a$, $b$, and $c$ are the coefficients of the quadratic equation $ax^2 + bx + c = 0$.
   For $x^2 + 3x - 28 = 0$, $a = 1$, $b = 3$, and $c = -28$.
   Plugging these values into the quadratic formula, we get $x = \frac{{-3 \pm \sqrt{{3^2 - 4(1)(-28)}}}}{{2(1)}}$.
   Simplifying further, we get $x = \frac{{-3 \pm \sqrt{{121}}}}{2}$.
   So, $x = \frac{{-3 + 11}}{2}$ or $x = \frac{{-3 - 11}}{2}$.
   This gives us $x = 4$ and $x = -7$.

Therefore, the solutions to the equation $x^2 + 3x - 28 = 0$ are $x = 4$ and $x = -7$.