Question

What are the approximate solutions of 2x2 + 9x = 8 to the nearest hundredth?

Answer 1

$2x^{2}$ + 9x = 8       Subtract 8 from both sides
$2x^{2}$ + 9x -8 = 0   Plug these into the Quadratic Formula

a = 2 , b = 9 , c = -8

x = $\frac-{b (+ or -) \sqrt{ b^{2} - 4ac} }{2a}$   Plug in the values
x = $\frac{-9 (+ or -) \sqrt{ 9^{2}-4(2)(-8) } }{2(2)}$   Multiply 2 and 2
x = $\frac{-9 (+ or -) \sqrt{ 9^{2}-4(2)(-8) } }{4}$   Multiply -4 and 2
x = $\frac{-9 (+ or -) \sqrt{ 9^{2}-8(-8) } }{4}$   Multiply -8 and -8
x = $\frac{-9 (+ or -) \sqrt{ 9^{2}+64} }{4}$   Square 9
x = $\frac{-9 (+ or -) \sqrt{81 + 64} }{4}$   Add 81 and 64
x = $\frac{-9 (+ or -) \sqrt{145} }{4}$ Take apart the (+ or -)
x = $\frac{-9 + \sqrt{145} }{4}$ and $\frac{-9 - \sqrt{145} }{4}$   Find the square root of 145
x ≈ $\frac{-9 +12.0415945788}{4}$ or $\frac{-9 -12.0415945788 }{4}$   Add -9 and the decimal , Subtract -9 and the decimal
x ≈ $\frac{3.0415945788}{4}$ or $\frac{-21.0415945788}{4}$   Divide both fractions
x ≈ 0.7603986447 or -5.2603986447   Round to the nearest hundredth (_._x)
x ≈ 0.76 or -5.26