Solve for x
x=\frac{\sqrt{1513}-27}{14}\approx 0.849807191
x=\frac{-\sqrt{1513}-27}{14}\approx -4.706950048
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7x^{2}+27x-28=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{-27±\sqrt{27^{2}-4\times 7\left(-28\right)}}{2\times 7}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 7 for a, 27 for b, and -28 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-27±\sqrt{729-4\times 7\left(-28\right)}}{2\times 7}
Square 27.
x=\frac{-27±\sqrt{729-28\left(-28\right)}}{2\times 7}
Multiply -4 times 7.
x=\frac{-27±\sqrt{729+784}}{2\times 7}
Multiply -28 times -28.
x=\frac{-27±\sqrt{1513}}{2\times 7}
Add 729 to 784.
x=\frac{-27±\sqrt{1513}}{14}
Multiply 2 times 7.
x=\frac{\sqrt{1513}-27}{14}
Now solve the equation x=\frac{-27±\sqrt{1513}}{14} when ± is plus. Add -27 to \sqrt{1513}.
x=\frac{-\sqrt{1513}-27}{14}
Now solve the equation x=\frac{-27±\sqrt{1513}}{14} when ± is minus. Subtract \sqrt{1513} from -27.
x=\frac{\sqrt{1513}-27}{14} x=\frac{-\sqrt{1513}-27}{14}
The equation is now solved.
7x^{2}+27x-28=0
Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.
7x^{2}+27x-28-\left(-28\right)=-\left(-28\right)
Add 28 to both sides of the equation.
7x^{2}+27x=-\left(-28\right)
Subtracting -28 from itself leaves 0.
7x^{2}+27x=28
Subtract -28 from 0.
\frac{7x^{2}+27x}{7}=\frac{28}{7}
Divide both sides by 7.
x^{2}+\frac{27}{7}x=\frac{28}{7}
Dividing by 7 undoes the multiplication by 7.
x^{2}+\frac{27}{7}x=4
Divide 28 by 7.
x^{2}+\frac{27}{7}x+\left(\frac{27}{14}\right)^{2}=4+\left(\frac{27}{14}\right)^{2}
Divide \frac{27}{7}, the coefficient of the x term, by 2 to get \frac{27}{14}. Then add the square of \frac{27}{14} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{27}{7}x+\frac{729}{196}=4+\frac{729}{196}
Square \frac{27}{14} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{27}{7}x+\frac{729}{196}=\frac{1513}{196}
Add 4 to \frac{729}{196}.
\left(x+\frac{27}{14}\right)^{2}=\frac{1513}{196}
Factor x^{2}+\frac{27}{7}x+\frac{729}{196}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.
\sqrt{\left(x+\frac{27}{14}\right)^{2}}=\sqrt{\frac{1513}{196}}
Take the square root of both sides of the equation.
x+\frac{27}{14}=\frac{\sqrt{1513}}{14} x+\frac{27}{14}=-\frac{\sqrt{1513}}{14}
Simplify.
x=\frac{\sqrt{1513}-27}{14} x=\frac{-\sqrt{1513}-27}{14}
Subtract \frac{27}{14} from both sides of the equation.
x ^ 2 +\frac{27}{7}x -4 = 0
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
r + s = -\frac{27}{7} rs = -4
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 = -\frac{27}{14} - u s = -\frac{27}{14} + u
Two numbers r and s sum up to -\frac{27}{7} exactly when the average of the two numbers is \frac{1}{2}*-\frac{27}{7} = -\frac{27}{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>
(-\frac{27}{14} - u) (-\frac{27}{14} + u) = -4
To solve for unknown quantity u, substitute these in the product equation rs = -4
\frac{729}{196} - u^2 = -4
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -4-\frac{729}{196} = -\frac{1513}{196}
Simplify the expression by subtracting \frac{729}{196} on both sides
u^2 = \frac{1513}{196} u = \pm\sqrt{\frac{1513}{196}} = \pm \frac{\sqrt{1513}}{14}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{27}{14} - \frac{\sqrt{1513}}{14} = -4.707 s = -\frac{27}{14} + \frac{\sqrt{1513}}{14} = 0.850
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
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