How do you solve #-10x^2 + 11x + 24 = 20# using the quadratic formula?

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Answer 1 · 2015-08-30T02:01:17

#x_(1,2) = (11 +- sqrt(281))/20#

For a general form quadratic equation

#color(blue)(ax^2 + bx + c = 0)#

you can use the quadratic formula to determine the roots of the equation

#color(blue)(x_(1,2) = (-b +- sqrt(b^2 - 4ac))/(2a))#

So, start by getting your equation into standard quadratic form. To do that, add #-20# to both sides of the equation

#-10x^2 + 11x + 24 - 20 = color(red)(cancel(color(black)(20))) - color(red)(cancel(color(black)(20)))#

#-10x^2 + 11x + 4 = 0#

In your case, you have #a = -10#, #b = 11#, and #c = 4#, which means that the quadratic formula will look like this

#x_(1,2) = (-11 +- sqrt(11^2 - 4 * (-10) * (4)))/(2 * (-10))#

#x_(1,2) = (-11 +- sqrt(281))/((-20)) = (11 +- sqrt(281))/20#

The two roots of the quadratic equation will thus be

#x_1 = (11 + sqrt(281))/(20)" "# and #" "x_2 = (11 - sqrt(281))/20#