Question

Which expression gives the solutions of -5+2x^2=-6x

Answer 1

-5 + 2x² = -6x
rearrange the equation to the form ax² + bx + c = 0

=>  2x² + 6x - 5

use the quadratic formula to solve for the value(s) of x  $-b ± \sqrt{ \frac{b^{2} - 4ac}{2a} }$

=>   $-6 ± \sqrt{ \frac{6^{2} - 4(2)(-5)}{2(2)} }$

=>   $-6 ± \sqrt{ \frac{36 - (-40)}{4} }$

=>  $-6 ± \sqrt{ \frac{76}{4} }$

∴  x =  $-6 + \sqrt{ 19} }$      OR   x = $-6 - \sqrt{19}$
       
          x =  - 1.64 ;  x = - 10.36

Answer 2

Answer:

The solutions are

$x1=\frac{-6+2\sqrt{19}}{4}$      

$x2=\frac{-6-2\sqrt{19}} {4}$

Step-by-step explanation:

we have

$-5+2x^{2} =-6x$

rewrite the quadratic equation

$2x^{2}+6x-5=0$

The formula to solve a quadratic equation of the form $ax^{2} +bx+c=0$ is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$2x^{2}+6x-5=0$

so

$a=2\\b=6\\c=-5$

substitute in the formula

$x=\frac{-6(+/-)\sqrt{6^{2}-4(2)(-5)}} {2(2)}$

$x=\frac{-6(+/-)\sqrt{76}} {4}$

$x=\frac{-6(+/-)2\sqrt{19}} {4}$

$x1=\frac{-6+2\sqrt{19}}{4}$      

$x2=\frac{-6-2\sqrt{19}} {4}$