When solving quadratic equations, the equation is set equal to zero, factorise, then each bracket is solved against zero.
Solve by factorising
\[x^2 + 5x + 6 = 0\]
\[(x + 2)(x + 3) = 0\]
Then, solve each factor:
\[x + 2 = 0 \quad \text{and} \quad x + 3 = 0\]
\[x = -2 \quad \text{and} \quad x = -3\]
Quadratics can either have 2, 1 or no solutions.
Using the Quadratic Formula
For any quadratic equation \(ax^2+bx+c=0\), the quadratic formula is:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
We can apply this formulae to any quadratic.
\[2x^2+3x−2=0\]
\[x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-2)}}{2(2)}\]
\[= \frac{-3 \pm \sqrt{9 + 16}}{4}\]
\[= \frac{-3 \pm \sqrt{25}}{4}\]
\[= \frac{-3 \pm 5}{4}\]
So the solutions are:
\[x = \frac{2}{4} = \frac{1}{2} \quad \text{or} \quad x = \frac{-8}{4} = -2\]