2 x+5x-4 = quadratic equation
Answer (1)
The value of x in [tex]2x^2 + 5x -4 =0[/tex] is [tex]x = \frac{-5 + \sqrt{57}}{4} \quad \text{and} \quad x = \frac{-5 - \sqrt{57}}{4}$$[/tex]To solve the equation, we must use the quadratic formula,[tex]\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]a=2b=5c=-4Substitute the values,[tex]x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 2 \cdot (-4)}}{2 \cdot 2}$$[/tex][tex]$$b^2 - 4ac = 5^2 - 4 \cdot 2 \cdot (-4) = 25 + 32 = 57$$[/tex]Let's put the discriminant back to the formula and solve the denominator[tex]x = \frac{-5 \pm \sqrt{57}}{4}$$[/tex]Therefore, the final solution to the quadratic equation is [tex]x = \frac{-5 + \sqrt{57}}{4} \quad \text{and} \quad x = \frac{-5 - \sqrt{57}}{4}$$[/tex]
