Solve 3x^2+2x-3

$$3 x ^ { 2 } + 2 x - 3$$

Factor

$$3\left(x-\frac{-\sqrt{10}-1}{3}\right)\left(x-\frac{\sqrt{10}-1}{3}\right)$$

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.

$$3x^{2}+2x-3=0$$

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.

$$x=\frac{-2±\sqrt{2^{2}-4\times 3\left(-3\right)}}{2\times 3}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\times 3\left(-3\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-2±\sqrt{4-12\left(-3\right)}}{2\times 3}$$

Multiply -12 times -3.

$$x=\frac{-2±\sqrt{4+36}}{2\times 3}$$

Add 4 to 36.

$$x=\frac{-2±\sqrt{40}}{2\times 3}$$

Take the square root of 40.

$$x=\frac{-2±2\sqrt{10}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-2±2\sqrt{10}}{6}$$

Now solve the equation x=\frac{-2±2\sqrt{10}}{6} when ± is plus. Add -2 to 2\sqrt{10}.

$$x=\frac{2\sqrt{10}-2}{6}$$

Divide -2+2\sqrt{10} by 6.

$$x=\frac{\sqrt{10}-1}{3}$$

Now solve the equation x=\frac{-2±2\sqrt{10}}{6} when ± is minus. Subtract 2\sqrt{10} from -2.

$$x=\frac{-2\sqrt{10}-2}{6}$$

Divide -2-2\sqrt{10} by 6.

$$x=\frac{-\sqrt{10}-1}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-1+\sqrt{10}}{3} for x_{1} and \frac{-1-\sqrt{10}}{3} for x_{2}.

$$3x^{2}+2x-3=3\left(x-\frac{\sqrt{10}-1}{3}\right)\left(x-\frac{-\sqrt{10}-1}{3}\right)$$

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 3

$$x ^ 2 +\frac{2}{3}x -1 = 0$$

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

$$r + s = -\frac{2}{3} rs = -1$$

Two numbers r and s sum up to -\frac{2}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{2}{3} = -\frac{1}{3}. 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>

$$r = -\frac{1}{3} - u s = -\frac{1}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -1

$$(-\frac{1}{3} - u) (-\frac{1}{3} + u) = -1$$

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

$$\frac{1}{9} - u^2 = -1$$

Simplify the expression by subtracting \frac{1}{9} on both sides

$$-u^2 = -1-\frac{1}{9} = -\frac{10}{9}$$

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

$$u^2 = \frac{10}{9} u = \pm\sqrt{\frac{10}{9}} = \pm \frac{\sqrt{10}}{3}$$

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

$$r =-\frac{1}{3} - \frac{\sqrt{10}}{3} = -1.387 s = -\frac{1}{3} + \frac{\sqrt{10}}{3} = 0.721$$

Evaluate

$$3x^{2}+2x-3$$