Solve \(2x^2+7x-15=0\).

A quadratic equation is solved by factorising whenever integer factors exist. Look for two numbers multiplying to \(a \times c = 2 \times (-15) = -30\) and adding to \(b = 7\); these are \(10\) and \(-3\). Splitting the middle term gives \(2x^2 + 10x - 3x - 15 = 2x(x + 5) - 3(x + 5)\), so

\((2x - 3)(x + 5) = 0\) [M1]

If a product of two factors is zero then at least one factor must be zero, which is the principle that turns one quadratic into two simple equations.

• \(2x - 3 = 0\) gives \(x = 1.5\) [A1]
• \(x + 5 = 0\) gives \(x = -5\) [A1]

Check both: \(2(1.5)^2 + 7(1.5) - 15 = 4.5 + 10.5 - 15 = 0\), and \(2(25) - 35 - 15 = 0\). The answer \(x = 1.5\) may also be written as \(\dfrac{3}{2}\). Reading the roots straight off the brackets without changing sign, giving \(x = 3\) and \(x = 5\), is the standard error: the root is the value that makes the bracket vanish, so \((x + 5)\) gives \(x = -5\).