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NEXUS

When solving quadratic equations, the equation is set equal to zero, factorise, then each bracket is solved against zero.

\[x^2 + 5x + 6 = 0\]

\[(x + 2)(x + 3) = 0\]

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.

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:

\[x = \frac{2}{4} = \frac{1}{2} \quad \text{or} \quad x = \frac{-8}{4} = -2\]

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