How do you solve #x^2-5x=-20# solve equation using quadratic formula?

Question

Abigail Allen · Accepted Answer

#x=(5+-isqrt(55))/2 #

Given - #x^2-5x=-20# #x^2-5x+20=0# #x=(-b+-sqrt(b^2-4ac))/(2a)# #x=(-(-5)+-sqrt((-5)^2-(4 xx 1 xx 20)))/(2 xx 1)# #x=(5+-sqrt((25-80)))/2 # #x=(5+-sqrt(-55))/2 # #x=(5+sqrt(-55))/2 # #x=(5-sqrt(-55))/2 # #x=(5+-isqrt(55))/2 #

Ethan Adkins · Answer

undefined To solve the equation $x^2 - 5x = -20$ using the quadratic formula, you first need to rearrange it into the standard form of a quadratic equation: $ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are constants.

In this case, we have:
$x^2 - 5x + 20 = 0$.

Now, identify the coefficients:
$a = 1$,
$b = -5$, and
$c = 20$.

Next, apply the quadratic formula:
$x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$.

Substitute the values of $a$, $b$, and $c$:
$x = \frac{{-(-5) \pm \sqrt{{(-5)^2 - 4(1)(20)}}}}{{2(1)}}$.

Now, simplify inside the square root:
$x = \frac{{5 \pm \sqrt{{25 - 80}}}}{{2}}$,
$x = \frac{{5 \pm \sqrt{{-55}}}}{{2}}$.

Since the discriminant ($b^2 - 4ac$) is negative, the solutions are complex. Simplify further:
$x = \frac{{5 \pm i\sqrt{{55}}}}{{2}}$.

Therefore, the solutions to the equation $x^2 - 5x = -20$ using the quadratic formula are:
$x = \frac{{5 + i\sqrt{{55}}}}{{2}}$ and $x = \frac{{5 - i\sqrt{{55}}}}{{2}}$.