Solve 30b^2+42b-24

$$30 b ^ { 2 } + 42 b - 24$$

Factor

$$30\left(b-\frac{-\sqrt{129}-7}{10}\right)\left(b-\frac{\sqrt{129}-7}{10}\right)$$

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.

$$30b^{2}+42b-24=0$$

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.

$$b=\frac{-42±\sqrt{42^{2}-4\times 30\left(-24\right)}}{2\times 30}$$

Square 42.

$$b=\frac{-42±\sqrt{1764-4\times 30\left(-24\right)}}{2\times 30}$$

Multiply -4 times 30.

$$b=\frac{-42±\sqrt{1764-120\left(-24\right)}}{2\times 30}$$

Multiply -120 times -24.

$$b=\frac{-42±\sqrt{1764+2880}}{2\times 30}$$

Add 1764 to 2880.

$$b=\frac{-42±\sqrt{4644}}{2\times 30}$$

Take the square root of 4644.

$$b=\frac{-42±6\sqrt{129}}{2\times 30}$$

Multiply 2 times 30.

$$b=\frac{-42±6\sqrt{129}}{60}$$

Now solve the equation b=\frac{-42±6\sqrt{129}}{60} when ± is plus. Add -42 to 6\sqrt{129}.

$$b=\frac{6\sqrt{129}-42}{60}$$

Divide -42+6\sqrt{129} by 60.

$$b=\frac{\sqrt{129}-7}{10}$$

Now solve the equation b=\frac{-42±6\sqrt{129}}{60} when ± is minus. Subtract 6\sqrt{129} from -42.

$$b=\frac{-6\sqrt{129}-42}{60}$$

Divide -42-6\sqrt{129} by 60.

$$b=\frac{-\sqrt{129}-7}{10}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-7+\sqrt{129}}{10} for x_{1} and \frac{-7-\sqrt{129}}{10} for x_{2}.

$$30b^{2}+42b-24=30\left(b-\frac{\sqrt{129}-7}{10}\right)\left(b-\frac{-\sqrt{129}-7}{10}\right)$$

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 30

$$x ^ 2 +\frac{7}{5}x -\frac{4}{5} = 0$$

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

$$r + s = -\frac{7}{5} rs = -\frac{4}{5}$$

Two numbers r and s sum up to -\frac{7}{5} exactly when the average of the two numbers is \frac{1}{2}*-\frac{7}{5} = -\frac{7}{10}. 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>

$$r = -\frac{7}{10} - u s = -\frac{7}{10} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{4}{5}

$$(-\frac{7}{10} - u) (-\frac{7}{10} + u) = -\frac{4}{5}$$

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

$$\frac{49}{100} - u^2 = -\frac{4}{5}$$

Simplify the expression by subtracting \frac{49}{100} on both sides

$$-u^2 = -\frac{4}{5}-\frac{49}{100} = -\frac{129}{100}$$

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

$$u^2 = \frac{129}{100} u = \pm\sqrt{\frac{129}{100}} = \pm \frac{\sqrt{129}}{10}$$

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

$$r =-\frac{7}{10} - \frac{\sqrt{129}}{10} = -1.836 s = -\frac{7}{10} + \frac{\sqrt{129}}{10} = 0.436$$

Evaluate

$$30b^{2}+42b-24$$