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4c^{2}-8c-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.
c=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 4\left(-3\right)}}{2\times 4}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -8 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
c=\frac{-\left(-8\right)±\sqrt{64-4\times 4\left(-3\right)}}{2\times 4}
Square -8.
c=\frac{-\left(-8\right)±\sqrt{64-16\left(-3\right)}}{2\times 4}
Multiply -4 times 4.
c=\frac{-\left(-8\right)±\sqrt{64+48}}{2\times 4}
Multiply -16 times -3.
c=\frac{-\left(-8\right)±\sqrt{112}}{2\times 4}
Add 64 to 48.
c=\frac{-\left(-8\right)±4\sqrt{7}}{2\times 4}
Take the square root of 112.
c=\frac{8±4\sqrt{7}}{2\times 4}
The opposite of -8 is 8.
c=\frac{8±4\sqrt{7}}{8}
Multiply 2 times 4.
c=\frac{4\sqrt{7}+8}{8}
Now solve the equation c=\frac{8±4\sqrt{7}}{8} when ± is plus. Add 8 to 4\sqrt{7}.
c=\frac{\sqrt{7}}{2}+1
Divide 8+4\sqrt{7} by 8.
c=\frac{8-4\sqrt{7}}{8}
Now solve the equation c=\frac{8±4\sqrt{7}}{8} when ± is minus. Subtract 4\sqrt{7} from 8.
c=-\frac{\sqrt{7}}{2}+1
Divide 8-4\sqrt{7} by 8.
c=\frac{\sqrt{7}}{2}+1 c=-\frac{\sqrt{7}}{2}+1
The equation is now solved.
4c^{2}-8c-3=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.
4c^{2}-8c-3-\left(-3\right)=-\left(-3\right)
Add 3 to both sides of the equation.
4c^{2}-8c=-\left(-3\right)
Subtracting -3 from itself leaves 0.
4c^{2}-8c=3
Subtract -3 from 0.
\frac{4c^{2}-8c}{4}=\frac{3}{4}
Divide both sides by 4.
c^{2}+\left(-\frac{8}{4}\right)c=\frac{3}{4}
Dividing by 4 undoes the multiplication by 4.
c^{2}-2c=\frac{3}{4}
Divide -8 by 4.
c^{2}-2c+1=\frac{3}{4}+1
Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
c^{2}-2c+1=\frac{7}{4}
Add \frac{3}{4} to 1.
\left(c-1\right)^{2}=\frac{7}{4}
Factor c^{2}-2c+1. 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(c-1\right)^{2}}=\sqrt{\frac{7}{4}}
Take the square root of both sides of the equation.
c-1=\frac{\sqrt{7}}{2} c-1=-\frac{\sqrt{7}}{2}
Simplify.
c=\frac{\sqrt{7}}{2}+1 c=-\frac{\sqrt{7}}{2}+1
Add 1 to both sides of the equation.
x ^ 2 -2x -\frac{3}{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 = 2 rs = -\frac{3}{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 = 1 - u s = 1 + u
Two numbers r and s sum up to 2 exactly when the average of the two numbers is \frac{1}{2}*2 = 1. 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>
(1 - u) (1 + u) = -\frac{3}{4}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{3}{4}
1 - u^2 = -\frac{3}{4}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{3}{4}-1 = -\frac{7}{4}
Simplify the expression by subtracting 1 on both sides
u^2 = \frac{7}{4} u = \pm\sqrt{\frac{7}{4}} = \pm \frac{\sqrt{7}}{2}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =1 - \frac{\sqrt{7}}{2} = -0.323 s = 1 + \frac{\sqrt{7}}{2} = 2.323
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.