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a+b=-3 ab=4\left(-7\right)=-28
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx-7. To find a and b, set up a system to be solved.
1,-28 2,-14 4,-7
Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -28.
1-28=-27 2-14=-12 4-7=-3
Calculate the sum for each pair.
a=-7 b=4
The solution is the pair that gives sum -3.
\left(4x^{2}-7x\right)+\left(4x-7\right)
Rewrite 4x^{2}-3x-7 as \left(4x^{2}-7x\right)+\left(4x-7\right).
x\left(4x-7\right)+4x-7
Factor out x in 4x^{2}-7x.
\left(4x-7\right)\left(x+1\right)
Factor out common term 4x-7 by using distributive property.
x=\frac{7}{4} x=-1
To find equation solutions, solve 4x-7=0 and x+1=0.
4x^{2}-3x-7=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(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 4\left(-7\right)}}{2\times 4}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -3 for b, and -7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-3\right)±\sqrt{9-4\times 4\left(-7\right)}}{2\times 4}
Square -3.
x=\frac{-\left(-3\right)±\sqrt{9-16\left(-7\right)}}{2\times 4}
Multiply -4 times 4.
x=\frac{-\left(-3\right)±\sqrt{9+112}}{2\times 4}
Multiply -16 times -7.
x=\frac{-\left(-3\right)±\sqrt{121}}{2\times 4}
Add 9 to 112.
x=\frac{-\left(-3\right)±11}{2\times 4}
Take the square root of 121.
x=\frac{3±11}{2\times 4}
The opposite of -3 is 3.
x=\frac{3±11}{8}
Multiply 2 times 4.
x=\frac{14}{8}
Now solve the equation x=\frac{3±11}{8} when ± is plus. Add 3 to 11.
x=\frac{7}{4}
Reduce the fraction \frac{14}{8} to lowest terms by extracting and canceling out 2.
x=-\frac{8}{8}
Now solve the equation x=\frac{3±11}{8} when ± is minus. Subtract 11 from 3.
x=-1
Divide -8 by 8.
x=\frac{7}{4} x=-1
The equation is now solved.
4x^{2}-3x-7=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}-3x-7-\left(-7\right)=-\left(-7\right)
Add 7 to both sides of the equation.
4x^{2}-3x=-\left(-7\right)
Subtracting -7 from itself leaves 0.
4x^{2}-3x=7
Subtract -7 from 0.
\frac{4x^{2}-3x}{4}=\frac{7}{4}
Divide both sides by 4.
x^{2}-\frac{3}{4}x=\frac{7}{4}
Dividing by 4 undoes the multiplication by 4.
x^{2}-\frac{3}{4}x+\left(-\frac{3}{8}\right)^{2}=\frac{7}{4}+\left(-\frac{3}{8}\right)^{2}
Divide -\frac{3}{4}, the coefficient of the x term, by 2 to get -\frac{3}{8}. Then add the square of -\frac{3}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{3}{4}x+\frac{9}{64}=\frac{7}{4}+\frac{9}{64}
Square -\frac{3}{8} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{3}{4}x+\frac{9}{64}=\frac{121}{64}
Add \frac{7}{4} to \frac{9}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{3}{8}\right)^{2}=\frac{121}{64}
Factor x^{2}-\frac{3}{4}x+\frac{9}{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{3}{8}\right)^{2}}=\sqrt{\frac{121}{64}}
Take the square root of both sides of the equation.
x-\frac{3}{8}=\frac{11}{8} x-\frac{3}{8}=-\frac{11}{8}
Simplify.
x=\frac{7}{4} x=-1
Add \frac{3}{8} to both sides of the equation.