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4x^{2}-15x-63=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{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\times 4\left(-63\right)}}{2\times 4}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -15 for b, and -63 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-15\right)±\sqrt{225-4\times 4\left(-63\right)}}{2\times 4}
Square -15.
x=\frac{-\left(-15\right)±\sqrt{225-16\left(-63\right)}}{2\times 4}
Multiply -4 times 4.
x=\frac{-\left(-15\right)±\sqrt{225+1008}}{2\times 4}
Multiply -16 times -63.
x=\frac{-\left(-15\right)±\sqrt{1233}}{2\times 4}
Add 225 to 1008.
x=\frac{-\left(-15\right)±3\sqrt{137}}{2\times 4}
Take the square root of 1233.
x=\frac{15±3\sqrt{137}}{2\times 4}
The opposite of -15 is 15.
x=\frac{15±3\sqrt{137}}{8}
Multiply 2 times 4.
x=\frac{3\sqrt{137}+15}{8}
Now solve the equation x=\frac{15±3\sqrt{137}}{8} when ± is plus. Add 15 to 3\sqrt{137}.
x=\frac{15-3\sqrt{137}}{8}
Now solve the equation x=\frac{15±3\sqrt{137}}{8} when ± is minus. Subtract 3\sqrt{137} from 15.
x=\frac{3\sqrt{137}+15}{8} x=\frac{15-3\sqrt{137}}{8}
The equation is now solved.
4x^{2}-15x-63=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.
4x^{2}-15x-63-\left(-63\right)=-\left(-63\right)
Add 63 to both sides of the equation.
4x^{2}-15x=-\left(-63\right)
Subtracting -63 from itself leaves 0.
4x^{2}-15x=63
Subtract -63 from 0.
\frac{4x^{2}-15x}{4}=\frac{63}{4}
Divide both sides by 4.
x^{2}-\frac{15}{4}x=\frac{63}{4}
Dividing by 4 undoes the multiplication by 4.
x^{2}-\frac{15}{4}x+\left(-\frac{15}{8}\right)^{2}=\frac{63}{4}+\left(-\frac{15}{8}\right)^{2}
Divide -\frac{15}{4}, the coefficient of the x term, by 2 to get -\frac{15}{8}. Then add the square of -\frac{15}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{15}{4}x+\frac{225}{64}=\frac{63}{4}+\frac{225}{64}
Square -\frac{15}{8} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{15}{4}x+\frac{225}{64}=\frac{1233}{64}
Add \frac{63}{4} to \frac{225}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{15}{8}\right)^{2}=\frac{1233}{64}
Factor x^{2}-\frac{15}{4}x+\frac{225}{64}. 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{15}{8}\right)^{2}}=\sqrt{\frac{1233}{64}}
Take the square root of both sides of the equation.
x-\frac{15}{8}=\frac{3\sqrt{137}}{8} x-\frac{15}{8}=-\frac{3\sqrt{137}}{8}
Simplify.
x=\frac{3\sqrt{137}+15}{8} x=\frac{15-3\sqrt{137}}{8}
Add \frac{15}{8} to both sides of the equation.
x ^ 2 -\frac{15}{4}x -\frac{63}{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 4
r + s = \frac{15}{4} rs = -\frac{63}{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{15}{8} - u s = \frac{15}{8} + u
Two numbers r and s sum up to \frac{15}{4} exactly when the average of the two numbers is \frac{1}{2}*\frac{15}{4} = \frac{15}{8}. 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{15}{8} - u) (\frac{15}{8} + u) = -\frac{63}{4}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{63}{4}
\frac{225}{64} - u^2 = -\frac{63}{4}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{63}{4}-\frac{225}{64} = -\frac{1233}{64}
Simplify the expression by subtracting \frac{225}{64} on both sides
u^2 = \frac{1233}{64} u = \pm\sqrt{\frac{1233}{64}} = \pm \frac{\sqrt{1233}}{8}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{15}{8} - \frac{\sqrt{1233}}{8} = -2.514 s = \frac{15}{8} + \frac{\sqrt{1233}}{8} = 6.264
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.