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4x^{2}+3x-2-8=0
Subtract 8 from both sides.
4x^{2}+3x-10=0
Subtract 8 from -2 to get -10.
a+b=3 ab=4\left(-10\right)=-40
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx-10. To find a and b, set up a system to be solved.
-1,40 -2,20 -4,10 -5,8
Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -40.
-1+40=39 -2+20=18 -4+10=6 -5+8=3
Calculate the sum for each pair.
a=-5 b=8
The solution is the pair that gives sum 3.
\left(4x^{2}-5x\right)+\left(8x-10\right)
Rewrite 4x^{2}+3x-10 as \left(4x^{2}-5x\right)+\left(8x-10\right).
x\left(4x-5\right)+2\left(4x-5\right)
Factor out x in the first and 2 in the second group.
\left(4x-5\right)\left(x+2\right)
Factor out common term 4x-5 by using distributive property.
x=\frac{5}{4} x=-2
To find equation solutions, solve 4x-5=0 and x+2=0.
4x^{2}+3x-2=8
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.
4x^{2}+3x-2-8=8-8
Subtract 8 from both sides of the equation.
4x^{2}+3x-2-8=0
Subtracting 8 from itself leaves 0.
4x^{2}+3x-10=0
Subtract 8 from -2.
x=\frac{-3±\sqrt{3^{2}-4\times 4\left(-10\right)}}{2\times 4}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 3 for b, and -10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-3±\sqrt{9-4\times 4\left(-10\right)}}{2\times 4}
Square 3.
x=\frac{-3±\sqrt{9-16\left(-10\right)}}{2\times 4}
Multiply -4 times 4.
x=\frac{-3±\sqrt{9+160}}{2\times 4}
Multiply -16 times -10.
x=\frac{-3±\sqrt{169}}{2\times 4}
Add 9 to 160.
x=\frac{-3±13}{2\times 4}
Take the square root of 169.
x=\frac{-3±13}{8}
Multiply 2 times 4.
x=\frac{10}{8}
Now solve the equation x=\frac{-3±13}{8} when ± is plus. Add -3 to 13.
x=\frac{5}{4}
Reduce the fraction \frac{10}{8} to lowest terms by extracting and canceling out 2.
x=-\frac{16}{8}
Now solve the equation x=\frac{-3±13}{8} when ± is minus. Subtract 13 from -3.
x=-2
Divide -16 by 8.
x=\frac{5}{4} x=-2
The equation is now solved.
4x^{2}+3x-2=8
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-2-\left(-2\right)=8-\left(-2\right)
Add 2 to both sides of the equation.
4x^{2}+3x=8-\left(-2\right)
Subtracting -2 from itself leaves 0.
4x^{2}+3x=10
Subtract -2 from 8.
\frac{4x^{2}+3x}{4}=\frac{10}{4}
Divide both sides by 4.
x^{2}+\frac{3}{4}x=\frac{10}{4}
Dividing by 4 undoes the multiplication by 4.
x^{2}+\frac{3}{4}x=\frac{5}{2}
Reduce the fraction \frac{10}{4} to lowest terms by extracting and canceling out 2.
x^{2}+\frac{3}{4}x+\left(\frac{3}{8}\right)^{2}=\frac{5}{2}+\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{5}{2}+\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{169}{64}
Add \frac{5}{2} 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{169}{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{169}{64}}
Take the square root of both sides of the equation.
x+\frac{3}{8}=\frac{13}{8} x+\frac{3}{8}=-\frac{13}{8}
Simplify.
x=\frac{5}{4} x=-2
Subtract \frac{3}{8} from both sides of the equation.