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
\(x=\frac{-6(+/-)\sqrt{6^{2}-4(2)(-5)}} {2(2)}\)
\(x=\frac{-6(+/-)\sqrt{76}} {4}\)
\(x=\frac{-6(+/-)2\sqrt{19}} {4}\)
\(x1=\frac{-6+2\sqrt{19}}{4}\)
\(x2=\frac{-6-2\sqrt{19}} {4}\)
