How do you solve #x^2+x-20=0#?

Question

Kennedy Adams · Accepted Answer

#x=4,-5#

#color(blue)(x^2+x-20=0#

Factor the equation first:

#rarrx^2+5x-4x-20=0#

#rarrx(x+5)-4(x+5)=0#

#color(purple)(rarr(x-4)(x+5)=0#

#~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~#

Next, solve by setting both equations to zero.

#rarrx-4=0#

Solve by adding four to each side:

#rarrxcancel(-4+4)=0+4#

#color(green)(rArrx=4#

#~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~#

#rarrx+5=0#

Subtract #5# from each side, and solve:

#rarrxcancel(+5-5)=0-5#

#color(green)(rArrx=-5#

#~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~#

#:.color(indigo)( ul bar |x=4, -5|#

Savannah Anderson · Answer

undefined To solve the equation $ x^2 + x - 20 = 0 $, you can use the quadratic formula, which is given by:

$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

In the equation $ ax^2 + bx + c = 0 $, identify $ a = 1 $, $ b = 1 $, and $ c = -20 $, then substitute these values into the quadratic formula and solve for $ x $:

$$ x = \frac{-1 \pm \sqrt{(1)^2 - 4(1)(-20)}}{2(1)} $$

$$ x = \frac{-1 \pm \sqrt{1 + 80}}{2} $$

$$ x = \frac{-1 \pm \sqrt{81}}{2} $$

$$ x = \frac{-1 \pm 9}{2} $$

This gives two possible solutions:

$$ x_1 = \frac{-1 + 9}{2} = 4 $$

$$ x_2 = \frac{-1 - 9}{2} = -5 $$

So, the solutions to the equation are $ x = 4 $ and $ x = -5 $.