Solve 10x^2+17x+4

$$10 x ^ { 2 } + 17 x + 4$$

Factor

$$10\left(x-\frac{-\sqrt{129}-17}{20}\right)\left(x-\frac{\sqrt{129}-17}{20}\right)$$

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.

$$10x^{2}+17x+4=0$$

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.

$$x=\frac{-17±\sqrt{17^{2}-4\times 10\times 4}}{2\times 10}$$

Square 17.

$$x=\frac{-17±\sqrt{289-4\times 10\times 4}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-17±\sqrt{289-40\times 4}}{2\times 10}$$

Multiply -40 times 4.

$$x=\frac{-17±\sqrt{289-160}}{2\times 10}$$

Add 289 to -160.

$$x=\frac{-17±\sqrt{129}}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{-17±\sqrt{129}}{20}$$

Now solve the equation x=\frac{-17±\sqrt{129}}{20} when ± is plus. Add -17 to \sqrt{129}.

$$x=\frac{\sqrt{129}-17}{20}$$

Now solve the equation x=\frac{-17±\sqrt{129}}{20} when ± is minus. Subtract \sqrt{129} from -17.

$$x=\frac{-\sqrt{129}-17}{20}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-17+\sqrt{129}}{20} for x_{1} and \frac{-17-\sqrt{129}}{20} for x_{2}.

$$10x^{2}+17x+4=10\left(x-\frac{\sqrt{129}-17}{20}\right)\left(x-\frac{-\sqrt{129}-17}{20}\right)$$

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 10

$$x ^ 2 +\frac{17}{10}x +\frac{2}{5} = 0$$

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

$$r + s = -\frac{17}{10} rs = \frac{2}{5}$$

Two numbers r and s sum up to -\frac{17}{10} exactly when the average of the two numbers is \frac{1}{2}*-\frac{17}{10} = -\frac{17}{20}. 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>

$$r = -\frac{17}{20} - u s = -\frac{17}{20} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{2}{5}

$$(-\frac{17}{20} - u) (-\frac{17}{20} + u) = \frac{2}{5}$$

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

$$\frac{289}{400} - u^2 = \frac{2}{5}$$

Simplify the expression by subtracting \frac{289}{400} on both sides

$$-u^2 = \frac{2}{5}-\frac{289}{400} = -\frac{129}{400}$$

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

$$u^2 = \frac{129}{400} u = \pm\sqrt{\frac{129}{400}} = \pm \frac{\sqrt{129}}{20}$$

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

$$r =-\frac{17}{20} - \frac{\sqrt{129}}{20} = -1.418 s = -\frac{17}{20} + \frac{\sqrt{129}}{20} = -0.282$$

Evaluate

$$10x^{2}+17x+4$$