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a+b=30 ab=-9\left(-16\right)=144
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -9x^{2}+ax+bx-16. To find a and b, set up a system to be solved.
1,144 2,72 3,48 4,36 6,24 8,18 9,16 12,12
Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 144.
1+144=145 2+72=74 3+48=51 4+36=40 6+24=30 8+18=26 9+16=25 12+12=24
Calculate the sum for each pair.
a=24 b=6
The solution is the pair that gives sum 30.
\left(-9x^{2}+24x\right)+\left(6x-16\right)
Rewrite -9x^{2}+30x-16 as \left(-9x^{2}+24x\right)+\left(6x-16\right).
-3x\left(3x-8\right)+2\left(3x-8\right)
Factor out -3x in the first and 2 in the second group.
\left(3x-8\right)\left(-3x+2\right)
Factor out common term 3x-8 by using distributive property.
x=\frac{8}{3} x=\frac{2}{3}
To find equation solutions, solve 3x-8=0 and -3x+2=0.
-9x^{2}+30x-16=0
All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.
x=\frac{-30±\sqrt{30^{2}-4\left(-9\right)\left(-16\right)}}{2\left(-9\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -9 for a, 30 for b, and -16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-30±\sqrt{900-4\left(-9\right)\left(-16\right)}}{2\left(-9\right)}
Square 30.
x=\frac{-30±\sqrt{900+36\left(-16\right)}}{2\left(-9\right)}
Multiply -4 times -9.
x=\frac{-30±\sqrt{900-576}}{2\left(-9\right)}
Multiply 36 times -16.
x=\frac{-30±\sqrt{324}}{2\left(-9\right)}
Add 900 to -576.
x=\frac{-30±18}{2\left(-9\right)}
Take the square root of 324.
x=\frac{-30±18}{-18}
Multiply 2 times -9.
x=-\frac{12}{-18}
Now solve the equation x=\frac{-30±18}{-18} when ± is plus. Add -30 to 18.
x=\frac{2}{3}
Reduce the fraction \frac{-12}{-18} to lowest terms by extracting and canceling out 6.
x=-\frac{48}{-18}
Now solve the equation x=\frac{-30±18}{-18} when ± is minus. Subtract 18 from -30.
x=\frac{8}{3}
Reduce the fraction \frac{-48}{-18} to lowest terms by extracting and canceling out 6.
x=\frac{2}{3} x=\frac{8}{3}
The equation is now solved.
-9x^{2}+30x-16=0
Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.
-9x^{2}+30x-16-\left(-16\right)=-\left(-16\right)
Add 16 to both sides of the equation.
-9x^{2}+30x=-\left(-16\right)
Subtracting -16 from itself leaves 0.
-9x^{2}+30x=16
Subtract -16 from 0.
\frac{-9x^{2}+30x}{-9}=\frac{16}{-9}
Divide both sides by -9.
x^{2}+\frac{30}{-9}x=\frac{16}{-9}
Dividing by -9 undoes the multiplication by -9.
x^{2}-\frac{10}{3}x=\frac{16}{-9}
Reduce the fraction \frac{30}{-9} to lowest terms by extracting and canceling out 3.
x^{2}-\frac{10}{3}x=-\frac{16}{9}
Divide 16 by -9.
x^{2}-\frac{10}{3}x+\left(-\frac{5}{3}\right)^{2}=-\frac{16}{9}+\left(-\frac{5}{3}\right)^{2}
Divide -\frac{10}{3}, the coefficient of the x term, by 2 to get -\frac{5}{3}. Then add the square of -\frac{5}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{10}{3}x+\frac{25}{9}=\frac{-16+25}{9}
Square -\frac{5}{3} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{10}{3}x+\frac{25}{9}=1
Add -\frac{16}{9} to \frac{25}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{5}{3}\right)^{2}=1
Factor x^{2}-\frac{10}{3}x+\frac{25}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.
\sqrt{\left(x-\frac{5}{3}\right)^{2}}=\sqrt{1}
Take the square root of both sides of the equation.
x-\frac{5}{3}=1 x-\frac{5}{3}=-1
Simplify.
x=\frac{8}{3} x=\frac{2}{3}
Add \frac{5}{3} to both sides of the equation.
x ^ 2 -\frac{10}{3}x +\frac{16}{9} = 0
Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.
r + s = \frac{10}{3} rs = \frac{16}{9}
Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C
r = \frac{5}{3} - u s = \frac{5}{3} + u
Two numbers r and s sum up to \frac{10}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{10}{3} = \frac{5}{3}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>
(\frac{5}{3} - u) (\frac{5}{3} + u) = \frac{16}{9}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{16}{9}
\frac{25}{9} - u^2 = \frac{16}{9}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{16}{9}-\frac{25}{9} = -1
Simplify the expression by subtracting \frac{25}{9} on both sides
u^2 = 1 u = \pm\sqrt{1} = \pm 1
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{5}{3} - 1 = 0.667 s = \frac{5}{3} + 1 = 2.667
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.