Write the following equation in standard form. Then solve. 6q^2+2q=5q^2-3q+24 The equation in standard form is square

To write the given equation in standard form, we need to move all terms to one side of the equation.<br /><br />Given equation: $6q^{2}+2q=5q^{2}-3q+24$<br /><br />Step 1: Move all terms to one side of the equation.<br />$6q^{2}+2q-(5q^{2}-3q+24)=0$<br />$6q^{2}+2q-5q^{2}+3q-24=0$<br />$q^{2}+5q-24=0$<br /><br />Therefore, the equation in standard form is $q^{2}+5q-24=0$.<br /><br />To solve the equation, we can use the quadratic formula:<br />$q=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$<br /><br />In this case, $a=1$, $b=5$, and $c=-24$.<br /><br />Substituting these values into the quadratic formula, we get:<br />$q=\frac{-5\pm\sqrt{5^{2}-4(1)(-24)}}{2(1)}$<br />$q=\frac{-5\pm\sqrt{25+96}}{2}$<br />$q=\frac{-5\pm\sqrt{121}}{2}$<br />$q=\frac{-5\pm11}{2}$<br /><br />Therefore, the solutions to the equation are:<br />$q_1=\frac{-5+11}{2}=\frac{6}{2}=3$<br />$q_2=\frac{-5-11}{2}=\frac{-16}{2}=-8$<br /><br />So, the solutions to the equation $6q^{2}+2q=5q^{2}-3q+24$ are $q=3$ and $q=-8$.