Solve 20x^2-133x+158

$$20 x ^ { 2 } - 133 x + 158$$

Factor

$$20\left(x-\frac{133-3\sqrt{561}}{40}\right)\left(x-\frac{3\sqrt{561}+133}{40}\right)$$

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.

$$20x^{2}-133x+158=0$$

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.

$$x=\frac{-\left(-133\right)±\sqrt{\left(-133\right)^{2}-4\times 20\times 158}}{2\times 20}$$

Square -133.

$$x=\frac{-\left(-133\right)±\sqrt{17689-4\times 20\times 158}}{2\times 20}$$

Multiply -4 times 20.

$$x=\frac{-\left(-133\right)±\sqrt{17689-80\times 158}}{2\times 20}$$

Multiply -80 times 158.

$$x=\frac{-\left(-133\right)±\sqrt{17689-12640}}{2\times 20}$$

Add 17689 to -12640.

$$x=\frac{-\left(-133\right)±\sqrt{5049}}{2\times 20}$$

Take the square root of 5049.

$$x=\frac{-\left(-133\right)±3\sqrt{561}}{2\times 20}$$

The opposite of -133 is 133.

$$x=\frac{133±3\sqrt{561}}{2\times 20}$$

Multiply 2 times 20.

$$x=\frac{133±3\sqrt{561}}{40}$$

Now solve the equation x=\frac{133±3\sqrt{561}}{40} when ± is plus. Add 133 to 3\sqrt{561}.

$$x=\frac{3\sqrt{561}+133}{40}$$

Now solve the equation x=\frac{133±3\sqrt{561}}{40} when ± is minus. Subtract 3\sqrt{561} from 133.

$$x=\frac{133-3\sqrt{561}}{40}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{133+3\sqrt{561}}{40} for x_{1} and \frac{133-3\sqrt{561}}{40} for x_{2}.

$$20x^{2}-133x+158=20\left(x-\frac{3\sqrt{561}+133}{40}\right)\left(x-\frac{133-3\sqrt{561}}{40}\right)$$

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 20

$$x ^ 2 -\frac{133}{20}x +\frac{79}{10} = 0$$

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

$$r + s = \frac{133}{20} rs = \frac{79}{10}$$

Two numbers r and s sum up to \frac{133}{20} exactly when the average of the two numbers is \frac{1}{2}*\frac{133}{20} = \frac{133}{40}. 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-gzdabgg4ehffg0hf.b01.azurefd.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{133}{40} - u s = \frac{133}{40} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{79}{10}

$$(\frac{133}{40} - u) (\frac{133}{40} + u) = \frac{79}{10}$$

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

$$\frac{17689}{1600} - u^2 = \frac{79}{10}$$

Simplify the expression by subtracting \frac{17689}{1600} on both sides

$$-u^2 = \frac{79}{10}-\frac{17689}{1600} = -\frac{5049}{1600}$$

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

$$u^2 = \frac{5049}{1600} u = \pm\sqrt{\frac{5049}{1600}} = \pm \frac{\sqrt{5049}}{40}$$

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

$$r =\frac{133}{40} - \frac{\sqrt{5049}}{40} = 1.549 s = \frac{133}{40} + \frac{\sqrt{5049}}{40} = 5.101$$

Evaluate

$$20x^{2}-133x+158$$