Question

Which of the following are solutions to the equation below?

Answer 1

B. -4 and C.-5 
3x^2+27x+60=0 divide by 3
3(x^2+9x+20)=0 simplify
3(x+4)(x+5)=0
Use the 0 property and check your work.

Answer 2

Answer:

The solutions are B. -4 and C. -5

Step-by-step explanation:

For a quadratic equation of the form $ax^2+bx+c=0$ the solutions are

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

For $\mathrm{}\quad a=3,\:b=27,\:c=60:\quad x_{1,\:2}=\frac{-27\pm \sqrt{27^2-4\cdot \:3\cdot \:60}}{2\cdot \:3}$

$x_1=\frac{-27+\sqrt{27^2-4\cdot \:3\cdot \:60}}{2\cdot \:3}\\\\x_1=\frac{-27+\sqrt{9}}{2\cdot \:3}\\\\x_1=\frac{-27+3}{2\cdot \:3}\\\\x_1=\frac{-24}{6} = -4$

$x_2=\frac{-27-\sqrt{27^2-4\cdot \:3\cdot \:60}}{2\cdot \:3}\\\\x_2=\frac{-27-\sqrt{9}}{2\cdot \:3}\\\\x_2=\frac{-27-3}{2\cdot \:3}\\\\x_2=-\frac{30}{6} = -5$