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4x^2+6x=32
We move all terms to the left:
4x^2+6x-(32)=0
a = 4; b = 6; c = -32;
Δ = b2-4ac
Δ = 62-4·4·(-32)
Δ = 548
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{548}=\sqrt{4*137}=\sqrt{4}*\sqrt{137}=2\sqrt{137}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{137}}{2*4}=\frac{-6-2\sqrt{137}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{137}}{2*4}=\frac{-6+2\sqrt{137}}{8} $
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