Answer:
[tex]\large\boxed{x=\dfrac{5-\sqrt{73}}{4}\ or\ x=\dfrac{5+\sqrt{73}}{4}}[/tex]
Step-by-step explanation:
[tex]\text{The quadratic formula for}\ ax^2+bx+c=0\\\\\text{If}\ b^2-4ac>0,\ \text{then the equation has two different solutiions:}\ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\text{If}\ b^2-4ac=0,\ \text{then the equation has one solution:}\ x=\dfrac{-b}{2a}\\\\\text{If}\ b^2-4ac<0,\ \text{then the equation has no solution.}[/tex]
[tex]\text{We have}\ 2x^2=5x+6.\ \text{Convert to the form of}\ ax^2+bx+c=0:\\\\2x^2=5x+6\qquad\text{subtract}\ 5x\ \text{and}\ 6\ \text{from both sides}\\\\2x^2-5x-6=0\\\\a=2,\ b=-5,\ c=-6\\\\b^2-4ac=(-5)^2-4(2)(-6)=25+48=73>0\\\\x=\dfrac{-(-5)\pm\sqrt{73}}{2(2)}=\dfrac{5\pm\sqrt{73}}{4}[/tex]
