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4t^{2}+16t+9=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.
t=\frac{-16±\sqrt{16^{2}-4\times 4\times 9}}{2\times 4}
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.
t=\frac{-16±\sqrt{256-4\times 4\times 9}}{2\times 4}
Square 16.
t=\frac{-16±\sqrt{256-16\times 9}}{2\times 4}
Multiply -4 times 4.
t=\frac{-16±\sqrt{256-144}}{2\times 4}
Multiply -16 times 9.
t=\frac{-16±\sqrt{112}}{2\times 4}
Add 256 to -144.
t=\frac{-16±4\sqrt{7}}{2\times 4}
Take the square root of 112.
t=\frac{-16±4\sqrt{7}}{8}
Multiply 2 times 4.
t=\frac{4\sqrt{7}-16}{8}
Now solve the equation t=\frac{-16±4\sqrt{7}}{8} when ± is plus. Add -16 to 4\sqrt{7}.
t=\frac{\sqrt{7}}{2}-2
Divide -16+4\sqrt{7} by 8.
t=\frac{-4\sqrt{7}-16}{8}
Now solve the equation t=\frac{-16±4\sqrt{7}}{8} when ± is minus. Subtract 4\sqrt{7} from -16.
t=-\frac{\sqrt{7}}{2}-2
Divide -16-4\sqrt{7} by 8.
4t^{2}+16t+9=4\left(t-\left(\frac{\sqrt{7}}{2}-2\right)\right)\left(t-\left(-\frac{\sqrt{7}}{2}-2\right)\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -2+\frac{\sqrt{7}}{2} for x_{1} and -2-\frac{\sqrt{7}}{2} for x_{2}.
x ^ 2 +4x +\frac{9}{4} = 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 4
r + s = -4 rs = \frac{9}{4}
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 = -2 - u s = -2 + u
Two numbers r and s sum up to -4 exactly when the average of the two numbers is \frac{1}{2}*-4 = -2. 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>
(-2 - u) (-2 + u) = \frac{9}{4}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{9}{4}
4 - u^2 = \frac{9}{4}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{9}{4}-4 = -\frac{7}{4}
Simplify the expression by subtracting 4 on both sides
u^2 = \frac{7}{4} u = \pm\sqrt{\frac{7}{4}} = \pm \frac{\sqrt{7}}{2}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-2 - \frac{\sqrt{7}}{2} = -3.323 s = -2 + \frac{\sqrt{7}}{2} = -0.677
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.