How do you solve $2 a^{2} = - 6 + 8 a$ $2a^2 = - 6+ 8a$ ?

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Answer 1 · 2018-03-23T01:13:24

$a = 1$ $a=1$ or $3$ $3$

$2 a^{2} = - 6 + 8 a$ $2a^2=-6+8a$

$2 a^{2} - 8 a + 6 = 0$ $2a^2-8a+6=0$

$a^{2} - 4 a + 3 = 0$ $a^2-4a+3=0$

$(a - 1) (a - 3) = 0$ $(a-1)(a-3)=0$

$a = 1$ $a=1$ or $3$ $3$

How do you solve 2a^2 = - 6+ 8a2a2=−6+8a2a^2 = - 6+ 8a?