Question

Solve the equation for x, where x is a real number (5 points): -3x^2 + 4x - 31 = 23

Answer 1

Subtract 23 to both sides so that the equation becomes -3x^2 + 4x - 54 = 0.
To find the solutions to this equation, we can apply the quadratic formula. This quadratic formula solves equations of the form ax^2 + bx + c = 0
                                          x = [ -b ± √(b^2 - 4ac) ] / (2a)
                                          x = [ -4 ± √(4^2 - 4(-3)(-54)) ] / ( 2(-3) )
                                          x = [ -4 ± √(16 - (648) ) ] / ( -6 )
                                          x = [ -4 ± √(-632) ] / ( -6)
Since √-632 is nonreal, the answer to this question is that there are no real solutions.

Answer 2

Do you realize that you have given out 50 points? 

Step One
Start by subtracting 23 from both sides.
-3x^2 + 4x - 31 -23 = 0
-3x^2 + 4x - 54 = 0

This does not give you a real result. We'll solve it anyway

Step two
bring out the quadratic equation and solve that
a = - 3
b = 4
c = -54

$\text{x = }\dfrac{ -b \pm \sqrt{b^{2} - 4ac } }{2a}$

$\text{x = }\dfrac{ -(4 ) \pm \sqrt{4^{2} - 4(-3)*(-54) } }{2(-3)}\\ \\\text{x = }\dfrac{ -(4) \pm \sqrt{16 - 648 } }{-6}\\ \\ \text{x = }\dfrac{ -(4) \pm \sqrt{-632 } }{-6}$
$\text{x = }\dfrac{ -(4) \pm \sqrt{4* -158 } }{-6}$
$\text{x = }\dfrac{ -(4) \pm 2\sqrt{ -158 } }{-6}$
$\text{x = }\dfrac{ -4 \pm 2\sqrt{ 158 }i}{-6}$x = 0.666 +/- 2 sqrt(158)i/ - 6x = 0.666 -/+ 4.1899ix1 = 0.666 - 4.1899ix2 = 0.666 + 4.1899i