Answer:
[tex]x=4.330,-5.403[/tex]
Step-by-step explanation:
The formula to solve a quadratic equation of the form
[tex]ax^{2} +bx+c=0[/tex]
is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]y=-2.096x^{2} -2.25x+9.038[/tex]
For y=-40
substitute
[tex]-40=-2.096x^{2} -2.25x+9.038[/tex]
[tex]-2.096x^{2} -2.25x+9.038+40=0[/tex]
[tex]-2.096x^{2} -2.25x+49.038=0[/tex]
so
[tex]a=-2.096\\b=-2.25\\c=49.038[/tex]
substitute in the formula
[tex]x=\frac{-(-2.25)(+/-)\sqrt{-2.25^{2}-4(-2.096)(49.038)}} {2(-2.096)}[/tex]
[tex]x=\frac{2.25(+/-)\sqrt{416.197}} {-4.192}[/tex]
[tex]x=\frac{2.25(+/-)20.401} {-4.192}[/tex]
[tex]x=\frac{2.25(+)20.401} {-4.192}=-5.403[/tex]
[tex]x=\frac{2.25(-)20.401} {-4.192}=4.330[/tex]
therefore
[tex]x=4.330,-5.403[/tex]
