Factor
9\left(z-\frac{-\sqrt{8665}-95}{18}\right)\left(z-\frac{\sqrt{8665}-95}{18}\right)
Evaluate
9z^{2}+95z+10
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9z^{2}+95z+10=0
Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.
z=\frac{-95±\sqrt{95^{2}-4\times 9\times 10}}{2\times 9}
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.
z=\frac{-95±\sqrt{9025-4\times 9\times 10}}{2\times 9}
Square 95.
z=\frac{-95±\sqrt{9025-36\times 10}}{2\times 9}
Multiply -4 times 9.
z=\frac{-95±\sqrt{9025-360}}{2\times 9}
Multiply -36 times 10.
z=\frac{-95±\sqrt{8665}}{2\times 9}
Add 9025 to -360.
z=\frac{-95±\sqrt{8665}}{18}
Multiply 2 times 9.
z=\frac{\sqrt{8665}-95}{18}
Now solve the equation z=\frac{-95±\sqrt{8665}}{18} when ± is plus. Add -95 to \sqrt{8665}.
z=\frac{-\sqrt{8665}-95}{18}
Now solve the equation z=\frac{-95±\sqrt{8665}}{18} when ± is minus. Subtract \sqrt{8665} from -95.
9z^{2}+95z+10=9\left(z-\frac{\sqrt{8665}-95}{18}\right)\left(z-\frac{-\sqrt{8665}-95}{18}\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-95+\sqrt{8665}}{18} for x_{1} and \frac{-95-\sqrt{8665}}{18} for x_{2}.
x ^ 2 +\frac{95}{9}x +\frac{10}{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.This is achieved by dividing both sides of the equation by 9
r + s = -\frac{95}{9} rs = \frac{10}{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{95}{18} - u s = -\frac{95}{18} + u
Two numbers r and s sum up to -\frac{95}{9} exactly when the average of the two numbers is \frac{1}{2}*-\frac{95}{9} = -\frac{95}{18}. 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{95}{18} - u) (-\frac{95}{18} + u) = \frac{10}{9}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{10}{9}
\frac{9025}{324} - u^2 = \frac{10}{9}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{10}{9}-\frac{9025}{324} = -\frac{8665}{324}
Simplify the expression by subtracting \frac{9025}{324} on both sides
u^2 = \frac{8665}{324} u = \pm\sqrt{\frac{8665}{324}} = \pm \frac{\sqrt{8665}}{18}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{95}{18} - \frac{\sqrt{8665}}{18} = -10.449 s = -\frac{95}{18} + \frac{\sqrt{8665}}{18} = -0.106
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
Examples
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{ x } ^ { 2 } - 4 x - 5 = 0
Trigonometry
4 \sin \theta \cos \theta = 2 \sin \theta
Linear equation
y = 3x + 4
Arithmetic
699 * 533
Matrix
\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]
Simultaneous equation
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
Limits
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}