Answer:
[tex]\displaystyle -5 \ and \ \frac{1}{2}[/tex]
Step-by-step explanation:
[tex]2x^2 + 9x =5[/tex]
we have to move 5 in the other side with a different sign, to get a 2 grade ecuation
[tex]2x^2 + 9x -5= 0[/tex]
now we know
a=2
b=9
c=-5
we have to find delta
[tex]\boxed{\Delta=b^2-4ac} \\\\ \Delta=9^2-4\cdot2\cdot(-5)\\ \Delta=81+40 \\ \Delta=121[/tex]
and now we have to use the formula to find the root square
[tex]\displaystyle \boxed{ X_{1,2}=\frac{-b\pm\sqrt{\Delta} }{2a} }[/tex]
[tex]\displaystyle X_1=\frac{-b-\sqrt{\Delta} }{2a} =\frac{-9-11}{2\cdot2}=\frac{-20}{4} =-5[/tex]
[tex]\displaystyle X_2=\frac{-b+\sqrt{\Delta} }{2a} =\frac{-9+\sqrt{121} }{2\cdot4} =\frac{-9+11}{4} =\frac{2}{4} =\frac{1}{2}[/tex]
2x² + 9x = 5
2x² + 9x - 5 = 0
2x² + 10x - x - 5 = 0
2x(x + 5) - (x + 5) = 0
(2x - 1)(x + 5) = 0
2x - 1 = 0 => 2x₁ = 1 = > x₁ = 1/2
x + 5 = 0 => x₂ = - 5
