Find the roots of the quadratic equation:
\$(x+3)(x+7) = 0 \$

by using this formula

\$ax^2+bx+c=0\$

After converting the equation

\$ x^2 + 3x + 7x + 21 = 0 \$

Then we get

\$ x^2 + 10x + 21 = 0 \$

by using the formula

\$ x = (-b \pm \sqrt (b^2 - 4ac))/(2a) \$

after converting the equation into formula

\$ x = (-(10) ± \sqrt((10)^2 - 4\times(1)\times(21)))/(2\times(1)) \$

finding the roots of the equation

\$ x = (-10 ± \sqrt(100 - 84))/2 \$

Subtract the numbers under the square root

\$ x = (-10 ± \sqrt(16))/2 \$

Take the multiples of '16'

\$ x = (-10 ± 4)/2\$

Taking common factor as [2] in the Numerator

\$ x = (2(-5 ± 2))/2 \$

calculating the terms

\$ x = (\cancel(2)^1(-5 ± 2))/\cancel(2)^1 \$

After Cancellation we get

\$ x = -5 ± 2 \$

the required terms are

(-5 + 2 , -5 - 2)

Answer

B