Solve 38t^2-3,403t+65,590

$$38 t ^ { 2 } - 3,403 t + 65,590$$

Factor

$$38\left(t-\frac{3403-\sqrt{1610729}}{76}\right)\left(t-\frac{\sqrt{1610729}+3403}{76}\right)$$

Steps Using the Quadratic Formula

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.

$$38t^{2}-3403t+65590=0$$

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.

$$t=\frac{-\left(-3403\right)±\sqrt{\left(-3403\right)^{2}-4\times 38\times 65590}}{2\times 38}$$

Square -3403.

$$t=\frac{-\left(-3403\right)±\sqrt{11580409-4\times 38\times 65590}}{2\times 38}$$

Multiply -4 times 38.

$$t=\frac{-\left(-3403\right)±\sqrt{11580409-152\times 65590}}{2\times 38}$$

Multiply -152 times 65590.

$$t=\frac{-\left(-3403\right)±\sqrt{11580409-9969680}}{2\times 38}$$

Add 11580409 to -9969680.

$$t=\frac{-\left(-3403\right)±\sqrt{1610729}}{2\times 38}$$

The opposite of -3403 is 3403.

$$t=\frac{3403±\sqrt{1610729}}{2\times 38}$$

Multiply 2 times 38.

$$t=\frac{3403±\sqrt{1610729}}{76}$$

Now solve the equation t=\frac{3403±\sqrt{1610729}}{76} when ± is plus. Add 3403 to \sqrt{1610729}.

$$t=\frac{\sqrt{1610729}+3403}{76}$$

Now solve the equation t=\frac{3403±\sqrt{1610729}}{76} when ± is minus. Subtract \sqrt{1610729} from 3403.

$$t=\frac{3403-\sqrt{1610729}}{76}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{3403+\sqrt{1610729}}{76} for x_{1} and \frac{3403-\sqrt{1610729}}{76} for x_{2}.

$$38t^{2}-3403t+65590=38\left(t-\frac{\sqrt{1610729}+3403}{76}\right)\left(t-\frac{3403-\sqrt{1610729}}{76}\right)$$

Steps Using Direct Factoring Method

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 38

$$x ^ 2 -\frac{3403}{38}x +\frac{32795}{19} = 0$$

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

$$r + s = \frac{3403}{38} rs = \frac{32795}{19}$$

Two numbers r and s sum up to \frac{3403}{38} exactly when the average of the two numbers is \frac{1}{2}*\frac{3403}{38} = \frac{3403}{76}. 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>

$$r = \frac{3403}{76} - u s = \frac{3403}{76} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{32795}{19}

$$(\frac{3403}{76} - u) (\frac{3403}{76} + u) = \frac{32795}{19}$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$-\frac{11580409}{5776} - u^2 = \frac{32795}{19}$$

Simplify the expression by subtracting -\frac{11580409}{5776} on both sides

$$-u^2 = \frac{32795}{19}--\frac{11580409}{5776} = -\frac{1610729}{5776}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{1610729}{5776} u = \pm\sqrt{\frac{1610729}{5776}} = \pm \frac{\sqrt{1610729}}{76}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =\frac{3403}{76} - \frac{\sqrt{1610729}}{76} = 28.077 s = \frac{3403}{76} + \frac{\sqrt{1610729}}{76} = 61.476$$

Evaluate

$$38t^{2}-3403t+65590$$