Question

Quadratic formula 2x²+9x-4=0

Answer 1

$2x^2+9x-4=0\ \ \ \ /:2\\\\x^2+\frac{9}{2}x-2=0\\\\x^2+2x\cdot\frac{9}{4}+(\frac{9}{4})^2-(\frac{9}{4})^2=2\\\\(x+\frac{9}{4})^2-\frac{81}{16}=2\\\\(x+\frac{9}{4})^2=\frac{32}{16}+\frac{81}{16}\\\\(x+\frac{9}{4})^2=\frac{113}{16}\iff x+\frac{9}{4}=-\sqrt\frac{113}{16}\ \vee\ x+\frac{9}{4}=\sqrt\frac{113}{16}\\\\x=-\frac{9}{4}-\frac{\sqrt{113}}{4}\ \vee\ x=-\frac{9}{4}+\frac{\sqrt{113}}{4}\\\\x=-\frac{9+\sqrt{113}}{4}\ \vee\ x=-\frac{9-\sqrt{113}}{4}$

Answer 2

$2x^2+9x-4=0 \\ \\a=2 , b =9 , \ c=- 4 \\ \\\Delta = b^{2}-4ac = 9^{2}-4*2* (- 4)= 81+32 =113 \\ \\x_{1}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{-9- \sqrt{113}}{4} \\\\x_{2}=\frac{-b+\sqrt{\Delta }}{2a} = \frac{-9+\sqrt{113}}{4}$