Solve 7y^2+5y-125

$$7 y ^ { 2 } + 5 y - 125$$

Factor

$$7\left(y-\frac{-5\sqrt{141}-5}{14}\right)\left(y-\frac{5\sqrt{141}-5}{14}\right)$$

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.

$$7y^{2}+5y-125=0$$

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.

$$y=\frac{-5±\sqrt{5^{2}-4\times 7\left(-125\right)}}{2\times 7}$$

Square 5.

$$y=\frac{-5±\sqrt{25-4\times 7\left(-125\right)}}{2\times 7}$$

Multiply -4 times 7.

$$y=\frac{-5±\sqrt{25-28\left(-125\right)}}{2\times 7}$$

Multiply -28 times -125.

$$y=\frac{-5±\sqrt{25+3500}}{2\times 7}$$

Add 25 to 3500.

$$y=\frac{-5±\sqrt{3525}}{2\times 7}$$

Take the square root of 3525.

$$y=\frac{-5±5\sqrt{141}}{2\times 7}$$

Multiply 2 times 7.

$$y=\frac{-5±5\sqrt{141}}{14}$$

Now solve the equation y=\frac{-5±5\sqrt{141}}{14} when ± is plus. Add -5 to 5\sqrt{141}.

$$y=\frac{5\sqrt{141}-5}{14}$$

Now solve the equation y=\frac{-5±5\sqrt{141}}{14} when ± is minus. Subtract 5\sqrt{141} from -5.

$$y=\frac{-5\sqrt{141}-5}{14}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-5+5\sqrt{141}}{14} for x_{1} and \frac{-5-5\sqrt{141}}{14} for x_{2}.

$$7y^{2}+5y-125=7\left(y-\frac{5\sqrt{141}-5}{14}\right)\left(y-\frac{-5\sqrt{141}-5}{14}\right)$$

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 7

$$x ^ 2 +\frac{5}{7}x -\frac{125}{7} = 0$$

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

$$r + s = -\frac{5}{7} rs = -\frac{125}{7}$$

Two numbers r and s sum up to -\frac{5}{7} exactly when the average of the two numbers is \frac{1}{2}*-\frac{5}{7} = -\frac{5}{14}. 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>

$$r = -\frac{5}{14} - u s = -\frac{5}{14} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{125}{7}

$$(-\frac{5}{14} - u) (-\frac{5}{14} + u) = -\frac{125}{7}$$

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

$$\frac{25}{196} - u^2 = -\frac{125}{7}$$

Simplify the expression by subtracting \frac{25}{196} on both sides

$$-u^2 = -\frac{125}{7}-\frac{25}{196} = -\frac{3525}{196}$$

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

$$u^2 = \frac{3525}{196} u = \pm\sqrt{\frac{3525}{196}} = \pm \frac{\sqrt{3525}}{14}$$

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

$$r =-\frac{5}{14} - \frac{\sqrt{3525}}{14} = -4.598 s = -\frac{5}{14} + \frac{\sqrt{3525}}{14} = 3.884$$

Evaluate

$$7y^{2}+5y-125$$