Questions on Algebra: Complex Numbers answered by real tutors!

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There are many ways to solve a quadratic polynomials For me, unless the problem requires a specific method OR unless the equation is simple enough to 'factor' in my head, I prefer to use the quadratic equation.
x = (-b +- sqrt( b^2-4*a*c ))/(2*a)

Just plug in -16 for a, 21 for b and 40 for c.
Another way to solve, if you have a graphing calculator, is to plot the graph and then see where the the plot crosses the x axis.
In either case, you get the same answers

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation ax^2+bx+c=0

(in our case -16x^2+21x+40 = 0

) has the following solutons:

$x[12] = (b+-sqrt( b^2-4ac ))/2\a$

For these solutions to exist, the discriminant b^2-4ac

should not be a negative number.

First, we need to compute the discriminant b^2-4ac

.

Discriminant d=3001 is greater than zero. That means that there are two solutions: $x[12] = (-21+-sqrt( 3001 ))/2\a$ .

$x[1] = (-(21)+sqrt( 3001 ))/2\-16 = -1.05566824060029$
$x[2] = (-(21)-sqrt( 3001 ))/2\-16 = 2.36816824060029$

Quadratic expression -16x^2+21x+40

can be factored:
-16x+21x+40 = -16(x--1.05566824060029)*(x-2.36816824060029)

Again, the answer is: -1.05566824060029, 2.36816824060029. Here's your graph:
graph( 500, 500, -10, 10, -20, 20, -16*x^2+21*x+40 )