Question

Solve the quadratic equation. $$x^{2}+12 x+20=0$$

Short answer

The solutions to the quadratic equation \(x^{2}+12x+20=0\) are \(x=-2\) and \(x=-10\).

Step-by-step solution

Identify the constants

The quadratic equation given is \(x^{2}+12x+20=0\). Here, the coefficients of \(x^{2}\), \(x\), and the constant term are 1, 12, and 20 respectively. So, \(a=1\), \(b=12\), and \(c=20\).

Write down the equation to be factored

The factored form of a quadratic equation \(ax^{2}+bx+c = 0\) looks like this: \((x-p)(x-q)=0\), where \(p\) and \(q\) are the roots of the equation. These roots can be found by looking for two numbers that multiply to be \(ac\) (or just \(c\) here since \(a=1\)), and add up to \(b\).

Factor the equation

Looking for two numbers that multiply to \(20\) and add to \(12\), we find \(2\) and \(10\). So, we can break down middle term \(12x\) into \(2x + 10x\). This leads us to rewrite the equation as \(x^{2} + 2x + 10x + 20 = 0\). We can then factor by grouping, to get \((x+2)(x+10)=0\).

Solve for x

Setting each group equal to zero gives the solutions: \(x+2=0\) gives \(x=-2\) and \(x+10=0\) gives \(x=-10\).